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Contents
PART II
Foreword
Preface
v
vii
7
...
1 Introduction
7
...
3 Methods of Integration
7
...
5 Integration by Partial Fractions
7
...
7 Definite Integral
7
...
9 Evaluation of Definite Integrals by Substitution
7
...
Application of Integrals
8
...
2 Area under Simple Curves
8
...
Differential Equations
9
...
2 Basic Concepts
9
...
4 Formation of a Differential Equation whose
General Solution is given
9
...
Vector Algebra
10
...
2 Some Basic Concepts
10
...
4 Addition of Vectors
385
391
424
424
424
427
429
xiv
10
...
6 Product of Two Vectors
432
441
11
...
1 Introduction
11
...
3 Equation of a Line in Space
11
...
5 Shortest Distance between Two Lines
11
...
7 Coplanarity of Two Lines
11
...
9 Distance of a Point from a Plane
11
...
Linear Programming
12
...
2 Linear Programming Problem and its Mathematical Formulation
12
...
Probability
13
...
2 Conditional Probability
13
...
4 Independent Events
13
...
6 Random Variables and its Probability Distributions
13
...
— JAMES B
...
1 Introduction
Differential Calculus is centred on the concept of the
derivative
...
Integral
Calculus is motivated by the problem of defining and
calculating the area of the region bounded by the graph of
the functions
...
e
...
W
...
Further, the formula that gives
all these anti derivatives is called the indefinite integral of the function and such
process of finding anti derivatives is called integration
...
For instance, if we know the instantaneous velocity of an
object at any instant, then there arises a natural question, i
...
, can we determine the
position of the object at any instant? There are several such practical and theoretical
situations where the process of integration is involved
...
These two problems lead to the two forms of the integrals, e
...
, indefinite and
definite integrals, which together constitute the Integral Calculus
...
The definite integral is also used to solve many interesting
problems from various disciplines like economics, finance and probability
...
7
...
Instead of differentiating a function,
we are given the derivative of a function and asked to find its primitive, i
...
, the original
function
...
Let us consider the following examples:
We know that
d
(sin x ) = cos x
dx
d x3
( ) = x2
dx 3
and
...
(2)
d x
(e ) = e x
...
We say
x3
and
3
ex are the anti derivatives (or integrals) of x2 and ex, respectively
...
Similarly, in (2) and (3),
d x
d
d x3
(e + C) ✂ e x
( + C) ✁ x 2 and
(sin x + C) cos x ,
dx
dx 3
dx
Thus, anti derivatives (or integrals) of the above cited functions are not unique
...
For this reason
C is customarily referred to as arbitrary constant
...
d
F (x) = f (x) , ✄ x ☎ I (interval),
More generally, if there is a function F such that
dx
then for any arbitrary real number C, (also called constant of integration)
d
✆ F (x) + C✝ = f (x), x ☎ I
dx
INTEGRALS
289
{F + C, C ✄ R} denotes a family of anti derivatives of f
...
To show this, let g and h
be two functions having the same derivatives on an interval I
...
e
...
In view of the above remark, it is justified to infer that the family {F + C, C ✄ R}
provides all possible anti derivatives of f
...
Symbolically, we write
Notation Given that
☎ f (x) dx = F (x) + C
...
For the sake of convenience, we mention below the following symbols/terms/phrases
with their meanings as given in the Table (7
...
Table 7
...
From these formulae, we can write down immediately the corresponding formulae
(referred to as standard formulae) for the integrals of these functions, as listed below
which will be used to find integrals of other functions
...
However, in any specific problem one has to keep it in mind
...
2
...
Then ✡ f (x) dx ✟ x ✠ C
...
But these integrals are very similar geometrically
...
By
assigning different values to C, we get different members of the family
...
In this case, each integral represents a parabola with
its axis along y-axis
...
The
curve y = x2 + 1 for C = 1 is obtained by shifting the parabola y = x2 one unit along
y-axis in positive direction
...
Thus, for each positive value of C,
each parabola of the family has its vertex on the positive side of the y-axis and for
negative values of C, each has its vertex along the negative side of the y-axis
...
1
...
In the Fig 7
...
The same is true when a < 0
...
,
dy
at these points equals 2a
...
Thus, ✡ 2 x dx ✟ x ✠ C ✟ FC (x) (say), implies that
292
MATHEMATICS
Fig 7
...
Further, the following equation (statement) ✂ f (x) dx F (x ) ✁ C y (say) ,
represents a family of curves
...
This is the geometrical interpretation of indefinite integral
...
2
...
(I) The process of differentiation and integration are inverses of each other in the
sense of the following results :
d
f (x) dx = f (x)
dx ☎
and
✂ f ✆(x) dx = f (x) + C, where C is any arbitrary constant
...
e
...
(II) Two indefinite integrals with the same derivative lead to the same family of
curves and so they are equivalent
...
f (x) dx and g (x) dx are equivalent
...
(1)
d
g (x ) dx
dx
= f (x) + g (x)
Thus, in view of Property (II), it follows by (1) and (2) that
f (x)
g (x) dx =
(IV) For any real number k,
d
k
dx
f (x) dx
k f (x) dx
Proof By the Property (I),
Also
and
f (x) dx = g (x) dx ,
☞✍
✏✒✔ ✔ ✑✓ ✔
✔
✞✕ ✗ ✖ ✞ ✗ ✞
✞ ✡✞
✔ ✘
✏✒ ✔ ✑✓ ✎
✞ ✡✞
✞✙ ✗ ✗ ✗ ✚
✞ ✗✞ ✗✗✞
Proof
☛✌✎
✄✂
✞
f (x) dx + C1 ,C1 R
R is customarily expressed by writing
without mentioning the parameter
...
(2)
g (x) dx
...
dx
f (x) dx = k
d
dx
f (x) dx = k f (x)
Therefore, using the Property (II), we have
k f (x) dx
k
f (x) dx
...
, fn and the real numbers, k1, k2,
...
kn f n (x) dx
= k1 f1 (x) dx k2 f 2 (x) dx
...
To find an anti derivative of a given function, we search intuitively for a function
whose derivative is the given function
...
We illustrate it
through some examples
...
Recall that
d
sin 2x = 2 cos 2x
dx
or
cos 2x =
d 1
1 d
✁
(sin 2x) =
✂ sin 2 x ✄
dx
2
2 dx
✆
✝
1
sin 2 x
...
Note that
Therefore, an anti derivative of cos 2x is
d 3
4
✞ x ✠ x ✟ = 3x2 + 4x3
...
(iii) We know that
d
1
d
1
1
(log x) ✡ , x ☛ 0 and [log ( – x)] ✡
( – 1) ✡ , x ☞ 0
dx
x
dx
–x
x
Combining above, we get
Therefore,
✑
d
1
✌ log x ✍ ✎ , x ✏ 0
dx
x
1
1
dx ✎ log x is one of the anti derivatives of
...
x
✠
1
+ C1 – C 2
x
Note From now onwards, we shall write only one constant of integration in the
final answer
...
Given that
F(0) = 3, which gives,
3 = 0 – 6 × 0 + C or C = 3
Hence, the required anti derivative is the unique function F defined by
F(x) = x4 – 6x + 3
...
Thus, if we know one anti derivative F of a function f, we can write
down an infinite number of anti derivatives of f by adding any constant to F
expressed by F(x) + C, C ✄ R
...
(ii) Sometimes, F is not expressible in terms of elementary functions viz
...
We are
therefore blocked for finding ✂ f (x) dx
...
For instance
✄
y 4 dy ✁
1 5
y4 1
y ✂C
✂C✁
4✂1
5
7
...
3 Comparison between differentiation and integration
1
...
2
...
e
...
3
...
Similarly, all functions
are not integrable
...
4
...
The integral of
a function is not so
...
e
...
5
...
When a polynomial function P is integrated,
the result is a polynomial whose degree is 1 more than that of P
...
We can speak of the derivative at a point
...
7
...
The derivative of a function has a geometrical meaning, namely, the slope of the
tangent to the corresponding curve at a point
...
8
...
Similarly,
the integral is used in calculating the distance traversed when the velocity at time
t is known
...
Differentiation is a process involving limits
...
7
...
The process of differentiation and integration are inverses of each other as
discussed in Section 7
...
2 (i)
...
1
Find an anti derivative (or integral) of the following functions by the method of inspection
...
sin 2x
2
...
e 2x
2
3x
4
...
sin 2x – 4 e
Find the following integrals in Exercises 6 to 20:
x 2 (1 –
1
) dx
x2
6
...
9
...
✂
☎
✝
12
...
☞
x3 ✡ x 2 ☛ x – 1
dx 14
...
2
x–
x3 ✟ 5 x 2 – 4
1 ✄
dx
dx
11
...
x ( 3 x 2 ✁ 2 x ✁ 3) dx
16
...
(2 x 2 – 3sin x ✁ 5 x ) dx
18
...
☞
20
...
cosec 2 x
cos 2 x
Choose the correct answer in Exercises 21 and 22
...
The anti derivative of ☎ x ✁
✝
(A)
1
3
1
x3
3
✌
1
2x 2
1 ✄
✆ equals
x✞
2
✌C
(B)
1
2 3 1 2
x ✌ x ✌C
3
2
3
1
2 2
3 2 1 2
x ✟ 2x 2 ✟ C
x ✟ x ✟C
(D)
3
2
2
d
3
f ( x) ✍ 4 x3 ✎ 4 such that f (2) = 0
...
If
dx
x
1 129
1 129
4
3
(A) x ✁ 3 ✎
(B) x ✁ 4 ✁
8
8
x
x
(C)
(C)
x4 ✁
1 129
✁
8
x3
(D)
x3 ✁
1 129
✎
8
x4
300
MATHEMATICS
7
...
It was based on inspection, i
...
, on the
search of a function F whose derivative is f which led us to the integral of f
...
Hence, we need to develop additional techniques or methods for finding the integrals
by reducing them into standard forms
...
Integration by Substitution
2
...
Integration by Parts
7
...
1 Integration by substitution
In this section, we consider the method of integration by substitution
...
Consider
I=
Put x = g(t) so that
We write
f (x) dx
dx
= g✂(t)
...
It is often important to guess what will be the useful
substitution
...
Example 5 Integrate the following functions w
...
t
...
Thus, we make the substitution
mx = t so that mdx = dt
...
Thus, we use the substitution x2 + 1 = t so that
2x dx = dt
...
Thus, we use the substitution
x is x 2 ✁
2
2 x
(iii) Derivative of
x ✄ t so that
1
2 x
dx ✄ dt giving dx = 2t dt
...
Thus, we use the substitution
1 ✠ x2
tan–1 x = t so that
dx
= dt
...
These will be used later without
reference
...
Then dx = dt
...
dx
cos x dx
(iii) ✌ 1 tan x ☛ ✌ cos x sin x
☞
☞
1 (cos x + sin x + cos x – sin x) dx
= 2✌
cos x ☞ sin x
303
304
MATHEMATICS
1
= 2 ✁ dx
1 cos x – sin x
dx
2 ✁ cos x sin x
x C1 1 cos x – sin x
= 2 ✂ 2 ✂ 2 ✄ cos x sin x dx
✂
...
2
Integrate the functions in Exercises 1 to 37:
1
...
4
...
ax ✑ b
✎ log x ✏
2
x
5
...
x x ✑ 2
1
9
...
x – x
1
12
...
x
9 – 4x2
3
...
x2
(2 ✒ 3 x 3 )3
16
...
x 1 ✟ 2 x 2
11
...
17
...
1 x2
19
...
21
...
sec2 (7 – 4x)
23
...
6cos x ✂ 4sin x
25
...
28
...
☎1 ✝ cos x ✆
1
33
...
tan x
sin x cos x
(x ✄ 1) ✠ x ✄ log x ✡
2
x
37
...
cot x log sin x
sin x
30
...
26
...
1 ✂ cot x
35
...
38
...
✏
sin 2 x cos 2 x
(A) tan x + cot x + C
(C) tan x cot x + C
(B) 10x + x10 + C
(D) log (10x + x10) + C
(B) tan x – cot x + C
(D) tan x – cot 2x + C
7
...
2 Integration using trigonometric identities
When the integrand involves some trigonometric functions, we use some known identities
to find the integral as illustrated through the following example
...
= – cos x
INTEGRALS
307
EXERCISE 7
...
sin2 (2x + 5)
2
...
cos 2x cos 4x cos 6x
4
...
sin3 x cos3 x
6
...
sin 4x sin 8x
10
...
cos 2 x – cos 2✂
cos x – cos ✂
16
...
1
sin x cos3 x
8
...
cos4 2x
9
...
sin 2 x
1 ✁ cos x
14
...
tan3 2x sec 2x
17
...
20
...
sin – 1 (cos x)
1
cos (x – a) cos (x – b)
Choose the correct answer in Exercises 23 and 24
...
sin 2 x ✝ cos 2 x
dx is equal to
23
...
✞
dx equals
cos 2 (e x x)
(A) – cot (exx) + C
(C) tan (ex) + C
(B) tan (xex) + C
(D) cot (ex) + C
7
...
5
...
Then dx = a sec2 ✌ d✌
...
Then dx = a sec ✌ tan ✌ d ✌
...
Then dx = a cos✌ d✌
...
Then dx = a sec2 ✥ d ✥
...
dx
(7) To find the integral ☎ 2
, we write
ax ✂ bx ✂ c
2
✆✞
b
c✝
b ✟ ✞ c b2 ✟✝
✆
2
ax + bx + c = a ✡ x ✂ x ✂ ☛ ✠ a ✡ ☞ x ✂ ✌ ✂ ☞ – 2 ✌ ☛
a
a✎
2a ✑ ✏ a 4a ✑ ✎☛
✍
✍✡✏
2
Now, put x ✒
c b2
b
2
✓ t so that dx = dt and writing – 2 ✔ ✕ k
...
(8) To find the integral of the type ☎
dx
ax ✂ bx ✂ c
obtain the integral using the standard formulae
...
A and B are thus obtained and hence the integral is reduced to
one of the known forms
...
Let us illustrate the above methods by some examples
...
4 (1)]
(i) We have ✁ 2
8
x✆4
x ✂ 16
x –4
(ii) ✡
dx
2x ✠ x
2
✟✡
dx
1 – ✝ x – 1✞
2
Put x – 1 = t
...
Therefore,
✁
dx
2x ✂ x
2
= ✁
dt
1 – t2
–1
= sin (t ) C
[by 7
...
Then dx = dt
dx
dt
1
t
–1
✁ x ☛ ✂ 6 x 13 = ✁ t 2 22 ✄ 2 tan 2 C
=
1
x–3
tan – 1
✏C
2
2
[by 7
...
4 (7)
...
Then dx = dt
...
4 (i)]
13 17
x☞ –
1
6 6 ☞C
log
=
1
13 17
17
x☞ ☞
6 6
=
1
6x ✛ 4
log
✜ C1
17
6 x ✜ 30
=
1
3x ✛ 2
1
1
✜ C1 ✜ log
log
17
x✜5
17
3
=
1
3x ✢ 2
1
1
log
✄ C , where C = C1 ✘ log
17
x✄5
17
3
INTEGRALS
(iii) We have
dx
✠
5x
dx
✁
☎
✠
2x
2x ✄
5 ✝✟
✂
5✆ x2 –
✞
1
5
=
Put x –
313
dx
✏
✡
☞
✍
2
1☛ ✡ 1☛
x – ✌ –☞ ✌
5✎ ✍ 5✎
(completing the square)
2
1
✑ t
...
5
dx
Therefore,
✔
5x
✒
✓
=
1
5
2x
=
=
1
5
dt
✛
✕1✖
✘
✙5✚
2
t2 – ✗
✜1✢
✥
✦5✧
log t ✣ t 2 – ✤
2
✣C
1
1
2x
log x – ★ x 2 –
5
5
5
★
[by 7
...
4 (9), we express
d
2
✩ 2 x ✫ 6 x ✫ 5 ✪ ✫ B = A (4 x ★ 6) ★ B
dx
Equating the coefficients of x and the constant terms from both sides, we get
x+2= A
1
1
and B =
...
(1)
314
MATHEMATICS
In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt
dt
I1 = ✂
t
Therefore,
log t ✁ C1
2
= log | 2 x ✄ 6 x ✄ 5 | ✄ C1
dx
dx
1
☎ ✝
I2 = ✝ 2
2 x ✆ 6 x ✆ 5 2 x 2 ✆ 3x ✆ 5
2
and
=
Put x ✍
...
4 (3)]
...
4 (10)
...
e
...
(1)
In I1, put 5 – 4x – x2 = t, so that (– 4 – 2x) dx = dt
...
(2)
dx
✌✞
9 – (x ✆ 2) 2
Put x + 2 = t, so that dx = dt
...
4 (5)]
x✎2
✎ C2
3
...
4
Integrate the functions in Exercises 1 to 23
...
4
...
x2 – 1
1 4x
2
3x
5
...
1
8
...
1
2
✒ 2 – x✓ ✔ 1
x2
6
...
sec 2 x
tan 2 x ✆ 4
316
MATHEMATICS
1
10
...
✁ x – 1✂✁ x – 2 ✂
4x ✝ 1
16
...
22
...
14
...
20
...
1
7 – 6x – x2
1
1
8 ✄ 3x – x
15
...
2
x –1
x 2
4x – x
☎ x – a ✆☎ x – b ✆
5x ✞ 2
1 2 x 3x 2
x 2
21
...
2
x
2
4 x 10
...
dx
equals
24
...
✡
dx
9x ✞ 4x2
(B) tan–1 (x + 1) + C
(D) tan–1x + C
equals
(A)
1 –1 ☞ 9 x ☛ 8 ✌
sin ✎
✏✍C
9
✑ 8 ✒
(B)
1
☞ 8x ☛ 9 ✌
sin –1 ✎
✏✍C
2
✑ 9 ✒
(C)
1 –1 ✔ 9 x ✓ 8 ✕
sin ✗
✘✖C
3
✙ 8 ✚
(D)
1
☞ 9x ☛ 8 ✌
sin –1 ✎
✏✍C
2
✑ 9 ✒
7
...
If the degree of P(x)
Q(x )
is less than the degree of Q(x), then the rational function is called proper, otherwise, it
is called improper
...
Thus, if
where T(x) is a polynomial in x and
P(x)
P(x)
is improper, then
Q(x )
Q(x)
T(x) ✁
317
P1 (x)
,
Q(x )
P1 (x)
is a proper rational function
...
The rational functions which we shall consider
here for integration purposes will be those whose denominators can be factorised into
P(x)
P(x)
dx , where
linear and quadratic factors
...
It is always possible to write the integrand as a sum of
simpler rational functions by a method called partial fraction decomposition
...
The following
Table 7
...
Table 7
...
No
...
px ✁ q
,a☎b
(x –a) (x –b)
A
B
✄
x–a x–b
2
...
px 2 ✟ qx ✟ r
(x – a ) (x – b) (x – c)
A
B
C
✄
✄
x–a x–b x–c
4
...
px 2 ✟ qx ✟ r
(x – a) (x 2 ✟ bx ✟ c)
A
Bx + C ,
✁ 2
x – a x ✁ bx ✁ c
where x2 + bx + c cannot be factorised further
In the above table, A, B and C are real numbers to be determined suitably
...
Therefore, by using the form of
partial fraction [Table 7
...
(1)
where, real numbers A and B are to be determined suitably
...
Equating the coefficients of x and the constant term, we get
A+B=0
and
2A + B = 1
Solving these equations, we get A =1 and B = – 1
...
e
...
Some authors use the symbol ‘✟’ to indicate that the statement is an
identity and use the symbol ‘=’ to indicate that the statement is an equation, i
...
, to
indicate that the statement is true only for certain values of x
...
Solving these equations, we get A = – 5 and B = 10
Let
Thus,
Therefore,
5
10
x2 ✄ 1
✂
= 1☎
x–2 x–3
x – 5x ✄ 6
2
1
dx
x2 1
✆ x 2 – 5 x 6 dx = ✝ dx ☎ 5 ✝ x – 2 dx ✂ 10✝ x – 3
= x – 5 log | x – 2 | + 10 log | x – 3 | + C
...
2 (4)
...
Solving these equations, we get
11
–5
–11
...
( x 2 ✂ 1) ( x 2 ✂ 4)
(x
2
x2
1) (x 2
4)
=
y
(y ✄ 1) (y ✄ 4)
y
A
B
☎
=
(y ☎ 1) (y ☎ 4)
y ☎1 y ☎ 4
Write
So that
y = A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give
A= ✆
Thus,
Therefore,
1
3
and B ✝
4
3
x2
1
4
☎
= –
2
2
2
(x ✂ 1) (x ✂ 4)
3 (x ☎ 1) 3 (x ☎ 4)
2
x 2 dx
dx
dx
1
4
✞ (x 2 ✂ 1) (x 2 ✂ 4) = – 3 ✠ x 2 ✟ 1 ✟ 3 ✠ x 2 ✟ 4
1 –1
4 1
–1 x
✡C
= – tan x ✡ ☛ tan
3
3 2
2
1 –1
2
–1 x
☞C
= – tan x ☞ tan
3
3
2
In the above example, the substitution was made only for the partial fraction part
and not for the integration part
...
✌ 3 sin ✎ – 2 ✍ cos ✎
d✎
Example 15 Find ✁
5 – cos 2✎ – 4 sin ✎
Solution Let y = sin ✏
Then
dy = cos✏ d✏
INTEGRALS
Therefore,
3 sin✂ – 2 ✁ cos✂
✄
5 – cos
2
✂
– 4 sin✂
d✂ =
☎
(3y – 2) dy
5 – (1 – y 2 ) – 4 y
=
☎
3y – 2
dy
y2 – 4 y ✆ 4
=
✠
=
A
B
✆
y ☞ 2 (y ☞ 2)2
3y – 2
3y – 2
Now, we write
✡
y – 2☛
2
✝
y – 2✞
2
✟
321
I (say)
[by Table 7
...
Therefore, the required integral is given by
I = ☎[
3
4
dy
dy
+
] dy = 3 ☎
+4☎
y – 2 (y – 2) 2
y–2
(y – 2)2
= 3 log y ✎ 2
✏
✓
= 3 log sin ✕ ☞ 2
✆
= 3 log (2 ☞ sin ✕) ✆
Example 16 Find
1
✌
4✑ –
y
✍
✒✏C
✎2✔
4
✆C
2 – sin ✕
4
+ C (since, 2 – sin ✖ is always positive)
2 ☞ sin ✕
x 2 ✏ x ✏ 1 dx
✗
(x ✏ 2) (x 2 ✏ 1)
Solution The integrand is a proper rational function
...
2(5)]
...
Solving these equations, we get
A
3
2
1
,B
and C
5
5
5
Thus, the integrand is given by
2
1
x✂
3
x2 ✁ x ✁ 1
3
1 2x ✂ 1
✂ 52 5 =
✂ ✄✆ 2 ☎✝
=
2
5 (x ✂ 2) x ✂ 1
5 (x ✂ 2) 5 ✞ x ✂ 1 ✟
(x ✁ 1) (x ✁ 2)
Therefore,
3 dx
1
2x
1
1
x2 ✁ x ✁ 1
✠ (x 2 +1) (x ✁ 2) dx = 5 ✡ x ✂ 2 ✂ 5 ✡ x 2 ✂ 1 dx ✂ 5 ✡ x 2 ✂ 1 dx
=
3
1
1
log x ☛ 2 ☛ log x 2 ☛ 1 ☛ tan –1 x ☛ C
5
5
5
EXERCISE 7
...
3
...
2
(x – 1) (x – 2) (x – 3)
x ✂ 3x ✂ 2
6
...
x
(x ☞ 1) (x – 1)
8
...
3x ☞ 5
x3 – x 2 ✌ x ☞ 1
10
...
5x
(x ☞ 1) (x 2 ✌ 4)
12
...
2
(1 ✌ x) (1 ☞ x 2 )
14
...
1
...
2
...
x (x n ☞ 1)
17
...
(x 2 1) (x 2
(x 2 3) (x 2
2)
4)
19
...
323
1
x (x 4 – 1)
1
[Hint : Put ex = t]
(e – 1)
Choose the correct answer in each of the Exercises 22 and 23
...
✄
equals
( x ✂ 1) ( x ✂ 2)
21
...
✄
equals
x ( x 2 ✍ 1)
1
2
(A) log x ✎ log (x +1) + C
2
1
(C) ✑ log x ✒ log (x 2 +1) + C
2
1
2
(B) log x ✏ log (x +1) + C
2
1
2
(D) log x ✏ log (x +1) + C
2
7
...
If u and v are any two differentiable functions of a single variable x (say)
...
(1)
✓ u dx dx = uv – ✓ v dx dx
dv
Let
u = f (x) and
= g (x)
...
e
...
Then, integration by parts gives
d
x cos x dx = x ✂ cos x dx – ✂ [ (x) ✂ cos x dx] dx
dx
= x sin x – sin x dx = x sin x + cos x + C
Suppose, we take
f (x) = cos x and g (x) = x
...
Therefore, the proper choice of the first
function and the second function is significant
...
For instance, the method does not work for
x sin x dx
...
(ii) Observe that while finding the integral of the second function, we did not add
any constant of integration
...
(iii) Usually, if any function is a power of x or a polynomial in x, then we take it as the
first function
...
Example 18 Find
log x dx
Solution To start with, we are unable to guess a function whose derivative is log x
...
Then,
the integral of the second function is x
...
1) dx = log x ☎ 1 dx ✄ ☎ [ (log x) ☎ 1 dx] dx
Hence,
dx
1
= (log x) ✆ x – ✟ x dx ✝ x log x – x ✞ C
...
The integral of the second
function is ex
...
Therefore,
x sin – 1 x
Example 20 Find ☛
1 ✡ x2
dx
Solution Let first function be sin – 1x and second function be
First we find the integral of the second function, i
...
,
Put t =1 – x2
...
...
x
Example 21 Find ✁ e sin x dx
Solution Take ex as the first function and sin x as second function
...
(1)
Taking e and cos x as the first and second functions, respectively, in I1, we get
x
I1 = e x sin x – ✁ e x sin x dx
Substituting the value of I1 in (1), we get
I = – ex cos x + ex sin x – I or 2I = ex (sin x – cos x)
ex
(sin x – cos x) + C
2
Alternatively, above integral can also be determined by taking sin x as the first function
and ex the second function
...
6
...
(1)
x
Taking f (x) and e as the first function and second function, respectively, in I1 and
x
integrating it by parts, we have I1 = f (x) ex – ✁ f ✏(x) e dx ✌ C
Substituting I1 in (1), we get
x
x
x
I = e f (x) ✁ f ✏(x) e dx ✌ ✁ e f ✏(x) dx ✌ C = ex f (x) + C
INTEGRALS
Thus,
✁
e x [ f ( x ) + f ( x )] dx = e x f ( x ) + C
Example 22 Find (i) ✁ e x (tan – 1 x ✂
1
) dx (ii)
1 ✂ x2
✄
(x 2 + 1) e x
dx
(x + 1)2
Solution
x
–1
(i) We have I = ✁ e (tan x ✂
1
) dx
1 ✂ x2
1
1 ✂ x2
Thus, the given integrand is of the form ex [ f (x) + f ☎(x)]
...
Consider f (x) ✌
Therefore,
x2 ✑ 1 x
x ✒1 x
e dx ✝
e ✑C
✄
2
x✑1
(x ✑ 1)
EXERCISE 7
...
1
...
x sin 3x
3
...
x sin– 1x
5
...
x2 log x
9
...
tan –1x
10
...
x (log x)2
11
...
(x2 + 1) log x
4
...
x tan–1 x
12
...
ex (sinx + cosx) 17
...
e ☎ 1 ✄ cos x ✆
✝
✞
1 ✠
x ✟1
19
...
e2x sin x
20
...
sin ✑
Choose the correct answer in Exercises 23 and 24
...
✖ x 2 e x dx equals
(A)
1 x3
e ✗C
3
(B)
1 x2
e ✘C
3
(C)
1 x3
e ✘C
2
(D)
1 x2
e ✘C
2
24
...
6
...
2
Example 23 Find ✝ x ☎ 2 x ☎ 5 dx
Solution Note that
2
2
✂ x ✞ 2 x ✞ 5 dx = ✂ (x ✞ 1) ✞ 4 dx
Put x + 1 = y, so that dx = dy
...
6
...
2
2
✁ 3 2x x dx = ✁ 4 y dy
Thus
=
y
1
4
y 4 ✂ y 2 ✄ sin –1 ✄ C
2
2
2
=
1
✆ x ☎ 1✝
(x ☎ 1) 3 ✞ 2 x ✞ x 2 ☎ 2 sin –1 ✟
✠☎C
2
✡ 2 ☛
[using 7
...
2 (iii)]
EXERCISE 7
...
1
...
1 ☞ 4x 2
3
...
x2 ✍ 4x ✍ 1
5
...
x2 ✍ 4x ✎ 5
7
...
x 2 ✌ 3x
9
...
10
...
3
2
x (1 ☎ x 2 ) 2 ☎ C
(C)
3
1
x2
1 ✕ x 2 ✕ x 2 log x ✕ 1 ✕ x 2 ✕ C
2
2
2
✁ x 8 x ✖ 7 dx is equal to
1
( x ✂ 4)
2
1
( x ✄ 4)
(B)
2
1
( x ✗ 4)
(C)
2
1
( x ✗ 4)
(D)
2
(A)
x 2 ✂ 8 x ✄ 7 ✄ 9log x ✂ 4 ✄ x 2 ✂ 8 x ✄ 7 ✄ C
x 2 ✂ 8 x ✄ 7 ✄ 9log x ✄ 4 ✄ x 2 ✂ 8 x ✄ 7 ✄ C
x 2 ✗ 8 x ✘ 7 ✗ 3 2 log x ✗ 4 ✘ x 2 ✗ 8 x ✘ 7 ✘ C
x2 ✗ 8x ✘ 7 ✗
9
log x ✗ 4 ✘ x 2 ✗ 8 x ✘ 7 ✘ C
2
INTEGRALS
331
7
...
In this
section, we shall study what is called definite integral of a function
...
A definite integral is denoted by
b
a
f (x) dx , where a is called the
lower limit of the integral and b is called the upper limit of the integral
...
e
...
Here, we shall consider these two cases separately as discussed
below:
7
...
1 Definite integral as the limit of a sum
Let f be a continuous function defined on close interval [a, b]
...
b
The definite integral ✁ f (x) dx is the area bounded by the curve y = f (x), the
a
ordinates x = a, x = b and the x-axis
...
2)
...
2
Divide the interval [a, b] into n equal subintervals denoted by [x0, x1], [x1, x2] ,
...
, [xn – 1, xn], where x0 = a, x1 = a + h, x2 = a + 2h,
...
We note that as n ✟ ☎✆ h ✟ 0
...
From Fig 7
...
(1)
Evidently as xr – xr–1 ✟ 0, i
...
, h ✟ 0 all the three areas shown in (1) become
nearly equal to each other
...
n 1
sn = h [f(x0) + … + f (xn - 1)] = h✂ f ( xr )
...
(3)
r ✁1
Here, sn and Sn denote the sum of areas of all lower rectangles and upper rectangles
raised over subintervals [xr–1, xr] for r = 1, 2, 3, …, n, respectively
...
(4)
As n ✝ ✞ strips become narrower and narrower, it is assumed that the limiting
values of (2) and (3) are the same in both cases and the common limiting value is the
required area under the curve
...
(5)
It follows that this area is also the limiting value of any area which is between that
of the rectangles below the curve and that of the rectangles above the curve
...
Thus, we rewrite (5) as
b
☛a
b
or
where
☞a
f ( x)dx = lim h [ f (a ) ✄ f (a ✄ h) ✄
...
✎ f (a ✎ (n – 1) h]
n
...
Remark The value of the definite integral of a function over any particular interval
depends on the function and the interval, but not on the variable of integration that we
INTEGRALS
333
choose to represent the independent variable
...
Hence, the variable of integration is called a dummy variable
...
Solution By definition
b
a
1
[ f (a ) ✆ f ( a ✆ h) ✆
...
✠ f (
)]
n
n
n
n
☞ (2n – 2) 2
✌
1
22
42
[1 ✍ ( 2 ✍ 1) ✍ ( 2 ✍ 1) ✍
...
✖ 1) ✖ ✓ (22 ✖ 42 ✖
...
✣ (n – 1)2 ]
n✜✢ n
n
= 2 lim
1
4 (n ✥ 1) n (2n – 1)
[n ✍ ✤
]
n✡☛ n
6
n
= 2 lim
1
2 (n ✦ 1) (2n – 1)
[n ✠
]
n✞✟ n
3
n
= 2 lim
= 2 lim [1 ✠
n✞✟
2
1
1
14
4
(1 ✦ ) (2 – )] = 2 [1 ✠ ] =
3
n
n
3
3
334
MATHEMATICS
Example 26 Evaluate
2 x
0
e dx as the limit of a sum
...
✝ e
✁ 0 e dx = (2 – 0) nlim
✂✄ n ✞
✠
2 x
Using the sum to n terms of a G
...
, where a = 1, r ☛
2n
2n – 2
n
2
en
✎
✆
✟
✡✟
, we have
✏
1 e n –1
1 ✑ e2– 1 ✒
2
lim
[
]
2
lim
e
dx
=
=
2
✁0
✒
n ✂✄ n
n✌✍ n ✑ 2
en☞1
✓ en– 1 ✔
2 x
=
2 (e 2 – 1)
✗ 2
✙ en – 1
lim ✙
2
n ✕✖
✙
✜ n
✘
✚
✚✛2
✚
✢
(e h ✤ 1)
✥ 1]
h✣ 0
h
[using lim
= e2 – 1
EXERCISE 7
...
1
...
b
a
4
✁1
5
x dx
2
...
✁ e dx
★1
1
3
2
3
...
✁ ( x ✦ e 2 x ) dx
0
7
...
8
...
Let x
be a given point in [a, b]
...
3
INTEGRALS
335
in Fig 7
...
The area of this shaded region depends upon
the value of x
...
We denote this
function of x by A(x)
...
(1)
Based on this definition, the two basic fundamental theorems have been given
...
7
...
2 First fundamental theorem of integral calculus
Theorem 1 Let f be a continuous function on the closed interval [a, b] and let A (x) be
the area function
...
7
...
3 Second fundamental theorem of integral calculus
We state below an important theorem which enables us to evaluate definite integrals
by making use of anti derivative
...
Then ✁ f ( x ) dx = [F( x )] ba = F (b) – F(a)
...
(ii) This theorem is very useful, because it gives us a method of calculating the
definite integral more easily, without calculating the limit of a sum
...
This strengthens the relationship
between differentiation and integration
...
3
1
For instance, the consideration of definite integral ✆ x( x 2 – 1) 2 dx is erroneous
☎2
1
2
since the function f expressed by f (x) = x( x – 1) 2 is not defined in a portion
– 1 < x < 1 of the closed interval [– 2, 3]
...
(i) Find the indefinite integral ✁ f ( x) dx
...
There is no need to keep
integration constant C because if we consider F(x) + C instead of F(x), we get
b
☎a
f ( x) dx ✂ [F ( x) ✄ C] ba ✂ [F(b) ✄ C] – [F(a) ✄ C] ✂ F(b) – F(a)
...
b
b
(ii) Evaluate F(b) – F(a) = [F ( x)] a , which is the value of ☎ f ( x) dx
...
Since ✠ x dx ✟
3
Therefore, by the second fundamental theorem, we get
I = F (3) – F (2) ✡
x
9
(ii) Let I ☛ ✁
4
(30 –
3
x 2 )2
dx
...
3
Put 30 – x 2 ☞ t
...
Consider ★ sin 2t cos 2 t dt
0
Put sin 2t = u so that 2 cos 2t dt = du or cos 2t dt =
So
★
sin 3 2t cos 2 t dt =
=
1
du
2
1 3
u du
2✩
1 4 1 4
[u ] ✪ sin 2t ✪ F (t ) say
8
8
Therefore, by the second fundamental theorem of integral calculus
1
1
✫
✫
I = F ( ) – F (0) ✪ [sin 4 – sin 4 0] ✪
4
8
2
8
338
MATHEMATICS
EXERCISE 7
...
1
...
31
✄2
☎
4
...
cosec x dx
9
...
2
✞0
4
☞ ☛
cos 2x dx
dx
1
✞0
1 – x2
6
✠
12
...
2
3
...
dx
2
✡0
3
5x2
17
...
(4 x 3 – 5 x 2
✟4
1
e dx
dx
x2
10
...
✞0
1
4
(2sec 2 x ✑ x3 ✑ 2) dx
dx
✠
5 x
18
...
✡0
11
...
✟0 x
dx
3
2x ✌ 3
dx
5x2 ✌ 1
✏
✒0
✁ 6 x ✁ 9)
1
✓
✄0
(sin 2
x
2
✍1
2
e x dx
x
x
– cos 2 ) dx
2
2
1
6x ✌ 3
✔x
x
) dx
dx
20
...
19
...
2
✞0
3
✞1
(A)
22
...
9 Evaluation of Definite Integrals by Substitution
In the previous sections, we have discussed several methods for finding the indefinite
integral
...
INTEGRALS
To evaluate
b
a
339
f ( x) dx , by substitution, the steps could be as follows:
1
...
2
...
3
...
4
...
Note In order to quicken this method, we can proceed as follows: After
performing steps 1, and 2, there is no need of step 3
...
✁
Let us illustrate this by examples
...
Solution Put t = x5 + 1, then dt = 5x4 dx
...
340
MATHEMATICS
Let
t = x5 + 1
...
Note that, when
x = – 1, t = 0 and when x = 1, t = 2
Thus, as x varies from – 1 to 1, t varies from 0 to 2
1
Therefore
✂ 1
5 x 4 x 5 ✁ 1 dx =
2
t dt
✂0
3 2
2 ✄ 2☎
✝t ✞
=
3 ✟✝ ✠✞
3
✆
0
3
☎
2✄ 2
2
4 2
2
(2 2) ✡
✝2 – 0 ✞ =
3 ✟✝
3
3
✞
✠
–1
tan x
dx
1 ☛ x2
1
Solution Let t = tan – 1x, then dt ✌
dx
...
Thus, as x varies from 0 to 1, t varies from 0 to
1
Therefore
☞0
tan –1 x
dx =
1 ☛ x2
✑
4
t
✘0
dt
✑
✒t2 ✓ 4
✔ ✕
✖ 2 ✗0
=
✎
4
...
10
Evaluate the integrals in Exercises 1 to 8 using substitution
...
✥0
2
4
...
2
☞0
sin ✧ cos5 ✧ d ✧ 3
...
1
dx
dx
7
...
✥0
8
...
1
☞0
sin x
dx
1 ✯ cos 2 x
2★ 1
☞1 ✪
✬x
–
1 ✩ 2x
✫ e dx
2 x2 ✭
1
9
...
If f (x) =
✂0
( x ✲ x3 ) 3
dx is
x4
(C) 3
t sin t dt , then f ✳(x) is
(A) cosx + x sin x
(C) x cosx
(B) x sinx
(D) sinx + x cosx
(D) 4
INTEGRALS
341
7
...
These will be useful in
evaluating the definite integrals more easily
...
In particular, ✄ a f ( x) dx ✂ 0
P2 :
✁ a f ( x) dx ✁ a f ( x) dx ☎ ✁ c f ( x) dx
P3 :
✄ a f ( x) dx ✂ ✄ a f (a ✆ b ✝ x) dx
P4 :
✄ 0 f ( x) dx ✂ ✄ 0 f (a ✝ x) dx
b
c
b
b
b
a
a
(Note that P4 is a particular case of P3)
2a
a
a
P5 :
✄ 0 f ( x) dx ✂ ✄ 0 f ( x) dx ✆ ✄ 0 f (2a ✝ x) dx
P6 :
✁ 0 f ( x) dx 2 ✁ 0 f ( x) dx, if f (2a ✞ x)
2a
a
f ( x) and
0 if f (2a – x) = – f (x)
P7 :
a
a
(i) ✁ f ( x) dx
✟a
2 ✁ f ( x) dx , if f is an even function, i
...
, if f (– x) = f (x)
...
e
...
✠a
We give the proofs of these properties one by one
...
Proof of P1 Let F be anti derivative of f
...
Proof of P2 Let F be anti derivative of f
...
(1)
c
✄ a f ( x) dx = F(c) – F(a)
...
(3)
342
MATHEMATICS
c
b
b
Adding (2) and (3), we get ✂ f ( x) dx ✂ f ( x) dx ✁ F(b) – F(a) ✁ ✂ f ( x) dx
a
c
a
This proves the property P2
...
Then dt = – dx
...
Therefore
b
a
✄ a f ( x) dx = ☎ ✄ b f (a ✆ b – t ) dt
b
= ✂ a f (a b – t ) dt (by P1)
b
= ✂ a f (a b – x) dx by P0
Proof of P4 Put t = a – x
...
When x = 0, t = a and when x = a, t = 0
...
2a
2a
a
Proof of P5 Using P2, we have ✂ f ( x) dx ✁ ✂ f ( x) dx ✂ f ( x) dx
...
Then
dt = – dx
...
Also x = 2a – t
...
(1)
f (2a – x) = f (x), then (1) becomes
2a
a
a
a
✂ 0 f ( x) dx = ✂ 0 f ( x) dx ✂ 0 f ( x) dx ✁ 2✂ 0 f ( x) dx,
and if
f (2a – x) = – f (x), then (1) becomes
2a
a
a
✂ 0 f ( x) dx = ✂ 0 f ( x) dx ✞✂ 0 f ( x) dx ✁ 0
Proof of P7 Using P2, we have
a
Let
0
a
✂ ✟ a f ( x) dx = ✄ ✠ a f ( x) dx ✆ ✄ 0 f ( x) dx
...
dt = – dx
...
Also x = – t
...
(1)
(i) Now, if f is an even function, then f (–x) = f (x) and so (1) becomes
a
✁ a
a
a
a
0
0
0
f ( x) dx ✄ ✁ f ( x ) dx ✂ ✁ f ( x) dx ✄ 2 ✁ f ( x) dx
(ii) If f is an odd function, then f (–x) = – f (x) and so (1) becomes
a
✁ a
a
a
0
0
f ( x) dx ✄ ☎ ✁ f ( x) dx ✂ ✁ f ( x) dx ✄ 0
Example 30 Evaluate
2
x3 – x dx
✁ 1
Solution We note that x3 – x ✡ 0 on [– 1, 0] and x3 – x
x3 – x ✡ 0 on [1, 2]
...
Therefore, by P7 (i), we get
✢
4
✣ –✢
4
✤
sin 2 x dx = 2 ✥ 4 sin 2 x dx
0
344
MATHEMATICS
=
2✂ 4
0
=
✞
x
✠
☛
(1 ✁ cos 2 x )
dx =
2
✄
4
✆0
✝
Example 32 Evaluate
Solution Let I =
✕
✗0
✕
✗0
✍✌
✑
✓4
–
1
1
✌✎
✌
sin ✒ – 0 ✏ –
2
2✔
4 2
x sin x
dx
1 ✖ cos 2 x
x sin x
dx
...
When x = 0, t = 1 and when x = ✛, t = – 1
...
Let f(x) = sin5 x cos4 x
...
e
...
Therefore, by P7 (ii), I = 0
INTEGRALS
Example 34 Evaluate
Solution Let I =
2
✂0
2
✂0
345
sin 4 x
dx
sin x ✁ cos 4 x
4
sin 4 x
dx
sin 4 x ✁ cos 4 x
...
(2)
Adding (1) and (2), we get
✟
2I =
Hence
I=
3
✏✍
✑
6
1 ✎ tan x
1✓
✑
I=
✗✖
3
✢
3
✤✕
✗✖
cos ✛
✢3
✑
✙
...
(2)
Adding (1) and (2), we get
2I =
✧
✧
3
✫✧
dx ✩ ✥ x ✦ ✧3
6
6
✩
★
3
✪
★
6
✩
★
6
...
Then 2 dx = dt, when x = 0, t = 0 and when x ✘
t = ✙
...
2
✗
2
,
INTEGRALS
347
EXERCISE 7
...
✂
1
...
2
✡0
7
...
cos5 x dx
sin 5 x ✠ cos5 x
sin x
2
☎0
sin x ✄ cos x
5
x (1 ✎ x) n dx
5
...
✁0
4
3
✆
dx 3
...
✌2
log (1 ✑ tan x) dx
9
...
2
✁0
(2log sin x ✒ log sin 2 x) dx
11
...
✕
x dx
✗0
1 ✖ sin x
✆
15
...
✓
2
✞0
4
✌0
13
...
✌0
17
...
✘
✌0
a
log (1 ☞ cos x) dx
x✝
x
dx
a✙ x
x ✍ 1 dx
19
...
✓
20
...
The value of
(A) 2
2
✫0
(C)
✢
(D) 1
✤
4 ✦ 3 sin x ✥
★ dx is
✩ 4 ✦ 3 cos x ✪
log ✧
(B)
3
4
(C) 0
(D) –2
348
MATHEMATICS
Miscellaneous Examples
Example 37 Find ✁ cos 6 x 1 sin 6 x dx
Solution Put t = 1 + sin 6x, so that dt = 6 cos 6x dx
1
Therefore
☎ cos 6 x 1 ✂ sin 6 x dx ✄
1 2
t dt
6☎
3
3
1 2 2
1
✆ (t ) ✝ C = (1 ✝ sin 6 x) 2 ✝ C
6 3
9
=
1
( x 4 ✞ x) 4
Example 38 Find ✟
dx
x5
Solution We have
Put 1 ☞
(x ✠
x5
4
☛
1
x) 4
1
dx ✡ ☛
1 4
)
x3 dx
x4
(1 ✠
1
3
✌ 1 – x – 3 ✌ t , so that 4 dx ✌ dt
3
x
x
1
5
1
1 4
( x 4 ✞ x) 4
1 4
Therefore ✟
dx
t
dt = ✑ t 4
✍
✟
5
3 5
3
x
5
4✎
1 ✏4
C = ✒1 ☞ 3 ✓
15 ✔
x ✕
C
x 4 dx
Example 39 Find ✘
( x ✖ 1) ( x 2 ✗ 1)
Solution We have
x4
= ( x 1)
( x ✙ 1) ( x 2 ✚ 1)
= ( x ✝ 1) ✝
Now express
1
x ☞x
3
2
x ☞1
1
( x ✛ 1) ( x 2 ✝ 1)
1
A
Bx ✂ C
✂ 2
=
2
( x ✜ 1) ( x ✂ 1)
( x ✛ 1)( x ✝ 1)
...
(2)
INTEGRALS
So
349
1 = A (x2 + 1) + (Bx + C) (x – 1)
= (A + B) x2 + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
1
1
, B C –
...
(3)
x
1
1
1
x4
✝
✝
= ( x ✆ 1) ✆
2
2
2
2( x ✝ 1) 2 ( x ✆ 1) 2( x ✆ 1)
( x ✄ 1) ( x ☎ x ☎ 1)
Therefore
x4
x2
1
1
1
2
–1
✡ ( x ✠ 1) ( x 2 ✟ x ✟ 1) dx ✞ 2 ✟ x ✟ 2 log x ✠ 1 – 4 log ( x ✟ 1) – 2 tan x ✟ C
☛
1 ☞
Example 40 Find ✒ ✍log (log x) ✌
✎ dx
(log x )2
✏
☛
Solution Let I ✓ ✒ ✍ log (log x) ✌
✏
✑
1 ☞
✎ dx
(log x)2 ✑
1
dx
(log x)2
In the first integral, let us take 1 as the second function
...
(1)
dx
Again, consider ✔
, take 1 as the second function and integrate it by parts,
log x
✖ x
✘
1
dx
✚ 1 ✛✙ ✗
we have ✮ log x ✜ ✢ log x – ✮ x ✤ – (log x) 2 ✦ x ✧ ✥ dx ✣
★ ✩ ✫ ✣✭
✢✬
✪
...
Then
✮
✯
I = 2✻
=
Example 42 Find ✽
Solution Let I ✒ ✔
dy
y ✺ ✰ 2✱
2
2
✹ 2 tan – 1
✲ t ✴ 1✳
✵ t✶
y
✸✺C
✺ C = 2 tan – 1 ✷
2
✕ t 2 ✼ 1✖
– 1 ✕ tan x ✼ 1 ✖
✗ 2 t ✘✘ ✓ C = 2 tan ✗✙ 2 tan x ✘✚ ✓ C
✙
✚
2 tan – 1 ✗
sin 2 x cos 2 x dx
9 – cos 4 (2 x)
sin 2 x cos 2 x
9 – cos 4 2 x
dx
2
INTEGRALS
351
Put cos2 (2x) = t so that 4 sin 2x cos 2x dx = – dt
Therefore
I✆–
1
4
dt
✏
Example 43 Evaluate
9–t
2
3
2
✆
–
1
1
✁t✂
sin –1 ✟ ✠ ✝ C ✆ ✞ sin
4
4
☞ 3✌
Solution Here f (x) = | x sin ✔x | =
Therefore
✡
✍3
☎
cos 2 2 x ☛ ✝ C
✎
x sin (✒ x) dx
✓ ✑1
3
2
|
✓ ✑1
1 ✄1
✘ x sin ✕ x for ✖ 1 ✗ x ✗ 1
✙
3
✚
✙ ✖ x sin ✕ x for 1 ✗ x ✗
2
✛
x sin ✒ x | dx =
1
✓ ✑1
x sin ✒ x dx
3
1
=
x sin ✒ x
3
dx ✜ ✓ 2 ✢
1
✦ ✣1
x sin ✤ x dx ✥ ✦ 2 x sin ✤ x dx
1
Integrating both integrals on righthand side, we get
3
2
|
✓ ✑1
x sin ✒ x | dx =
✧–
✪
✬
Solution Let I =
Thus
✶
✓0
✵
✦0
✤
1
✩
sin ✤ x ★
✤
2
✧ ✥ x cos ✤
✥
✫
✪
✤
✭✣ 1 ✬
3
x
✩
sin ✤ x ★ 2
✤
2
✫
✭1
2
1✯ 3 1
✮ 1
✢ ✢
✢
✰
✜
2
2
✱
✲
✒ ✳ ✒
✒✴
✒
✒
=
Example 44 Evaluate
x cos ✤ x
x dx
a cos x ✩ b 2 sin 2 x
2
2
✶
x dx
(✒ ✢ x) dx
✰
(using P4)
✓0 2
2
2
2
a cos x ✜ b sin x
a cos (✒ ✢ x) ✜ b2 sin 2 (✒ ✢ x)
2
2
=
✵
✤✦
0
✵
dx
x dx
✥✦
2
2
2
2
0
a cos x ✩ b sin x
a cos x ✩ b 2 sin 2 x
=
✵
✤ ✦
0
dx
✥I
a 2 cos 2 x ✩ b 2 sin 2 x
2I =
✵
✤✦
0
dx
a cos x ✩ b 2 sin 2 x
2
2
2
2
352
MATHEMATICS
or
dx
a cos x ☎ b 2 sin 2 x
✁
I=
2
✆0
2
2
✂
✁
2
✄2✆
2
0
dx
a cos x ☎ b 2 sin 2 x
2
2
(using P6)
✝
=
✞
sec2 x dx
a 2 ✟ b 2 tan 2 x
2
✠0
(dividing numerator and denominator by cos2 x)
...
Also, when x = 0, t = 0, and when x ☛
t
✡
2
,
☞ ✌
...
Miscellaneous Exercise on Chapter 7
Integrate the functions in Exercises 1 to 24
...
4
...
9
...
1
x ✛ x3
2
...
cos 3 x elog sinx
5
...
20
...
x ax ✢ x
1
1
x2 ★
1
x3
8
...
sin 8 ✯ cos8 x
1 ✯ 2sin 2 x cos 2 x
11
...
ex
(1 ✰ e x ) (2 ✰ e x )
14
...
e3 logx (x4 + 1)– 1
17
...
1
19
...
1 ✹ cos 2 x
22
...
tan
24
...
–1
✕
✒
✟ x ✡ 1 ✠ sin x ☛
cos 2 x dx
4 sin x cos x
2
e
dx
25
...
✔
27
...
✢ 0
28
...
✗ 0
9 ✖ 16 sin 2 x
1✛ x ✜ x
sin 2 x
6
✣
✦ x tan x
dx
32
...
✥ 2 sin 2 x tan ✤1 (sin x) dx
0
4
33
...
✮ 1 2
3
x ( x ✭ 1) 3
1
x
35
...
✫ ✰1 x cos x dx ✯ 0
37
...
Evaluate ✫ e 2✰ 3 x dx as a limit of a sum
...
1
(B) tan–1 (e–x) + C
(D) log (ex + e–x) + C
cos 2 x
dx is equal to
42
...
✸ sin ✴ x dx ✶ ✷ 1
0
2
38
...
✗ x
e ✖ e✹ x
(A) tan–1 (ex) + C
(C) log (ex – e–x) + C
2
3
2
✢ 0 sin x dx ✲ 3
(B) log |sin x ✖ cos x | ✖ C
(D)
1
(sin x ✁ cos x)2
353
354
MATHEMATICS
43
...
The value of ✎ tan ✝1 ✡
dx is
2 ☛
0
✌ 1☞ x ✞ x ✍
(A) 1
(B) 0
(C) –1
(D)
✏
4
Summary
Integration is the inverse process of differentiation
...
Thus, integration is a process which is the inverse of
differentiation
...
Then we write ✕ f ( x) dx ✓ F ( x) ✔ C
...
All these integrals differ by a constant
...
✑ Some properties of indefinite integrals are as follows:
✑
1
...
For any real number k, ✕ k f ( x) dx ✓ k ✕ f ( x) dx
More generally, if f1, f2, f3,
...
,kn are real
numbers
...
✔ kn f n ( x )] dx
= k1 ✕ f1 ( x) dx ✔ k2 ✕ f 2 ( x) dx ✔
...
Particularly, ✟ dx ✝ x ✞ C
☎
n ✄1
(iii) ✟ sin x dx ✝ – cos x ✞ C
✟ cos x dx ✝ sin x ✞ C
(v) ✟ cosec x dx ✝ – cot x ✞ C
✟ sec x dx ✝ tan x ✞ C
✟ sec x tan x dx ✝ sec x ✞ C
dx
✝ sin ✠ x ✞ C
cosec
x
cot
x
dx
–
cosec
x
C
✝
✞
✟
(viii)
✟
1✡ x
dx
dx
✝ tan ✠ x ✞ C
✟ 1 ✡ x ✝ ✡ cos✠ x ✞ C
(x) ✟
1✞ x
n 1
(i)
(ii)
(iv)
(vi)
(vii)
(ix)
x n dx
2
2
1
2
1
1
2
2
dx
✎ 1 ✍ x ☞ ✌ cot ☛ x ✍ C
(xii)
(xiii)
✕ a dx ✓ log a ✔ C
(xiv)
(xv)
✟ x x ✡ 1 ✝ ✡ cosec✠ x ✞ C
(xi)
1
2
ax
x
dx
1
2
(xvi)
✒ e dx ✏ e ✑ C
x
x
dx
✟ x x ✡ 1 ✝ sec✠ x ✞ C
1
✘ x dx ✖ log | x | ✗ C
1
2
Integration by partial fractions
P( x)
,
Q( x)
where P(x) and Q (x) are polynomials in x and Q (x) 0
...
T(x)
being polynomial can be easily integrated
...
A
2
...
px 2 ✆ qx ✆ r
( x ✝ a ) ( x ✝ b) ( x ✝ c)
=
4
...
px 2 ✆ qx ✆ r
( x ✝ a) ( x 2 ✆ bx ✆ c)
=
B
( x ✁ a)2
x✁a
A
x✄a
✂
A
x✟a
x✄b
✂
B
( x ✁ a) 2
x✁a
A
B
✞
C
x✄c
C
x✁b
Bx + C
x ✞ bx ✞ c
2
2
where x + bx + c can not be factorised further
...
The method in which we change the variable to some
other variable is called the method of substitution
...
Using substitution technique, we obtain the following standard
integrals
...
e
...
Care must be taken in choosing the first
function and the second function
...
x
x
✆ ✄ e [ f ( x) ✁ f ✏( x)] dx ✄ e f ( x) dx ✁ C
✆ Some special types of integrals
x 2
a2
2
2
x ✑ a 2 ✑ log x ✓ x 2 ✑ a 2 ✓ C
(i) ✔ x ✑ a dx ✒
2
2
x 2
a2
2
2
x ✓ a 2 ✓ log x ✓ x 2 ✓ a 2 ✓ C
(ii) ✔ x ✓ a dx ✒
2
2
x 2
a2
x
2
2
a ✑ x2 ✓
sin ✕1 ✓ C
(iii) ✔ a ✑ x dx ✒
2
2
a
dx
dx
or ✄
(iv) Integrals of the types ✄ 2
can be
2
ax ✁ bx ✁ c
ax ✁ bx ✁ c
transformed into standard form by expressing
c✗
✖ 2 b
ax2 + bx + c = a ✚ x ✁ x ✁ ✛
a
a✤
✣
✖✘
✣✚✥
a ✚✜ x ✁
b ✙ ✘ c b2 ✙✗
✁✜ ✂
✢✛
2a ✦✢ ✥ a 4a 2 ✦ ✤✛
2
px ✁ q dx
px ✁ q dx
(v) Integrals of the types ✄ ax 2 ✁ bx ✁ c or ✄
can be
ax 2 ✁ bx ✁ c
358
MATHEMATICS
transformed into standard form by expressing
d
(ax 2 bx c) B ✁ A (2ax b) B , where A and B are
dx
determined by comparing coefficients on both sides
...
Let x be a
x
given point in [a, b]
...
This concept of area function leads to the Fundamental Theorems of Integral
Calculus
...
Then A✆ (x) = f (x) for all
x ✝ [a, b]
...
a
This is called the definite integral of f over the range [a, b], where a and b
are called the limits of integration, a being the lower limit and b the
upper limit
...
– BIRKHOFF
8
...
Such formulae are
fundamental in the applications of mathematics to many
real life problems
...
However, they are inadequate for calculating the areas
enclosed by curves
...
In the previous chapter, we have studied to find the
area bounded by the curve y = f (x), the ordinates x = a,
x = b and x-axis, while calculating definite integral as the
limit of a sum
...
We shall also deal with finding
the area bounded by the above said curves
...
L
...
2 Area under Simple Curves
In the previous chapter, we have studied
definite integral as the limit of a sum and
how to evaluate definite integral using
Fundamental Theorem of Calculus
...
From Fig 8
...
Consider
an arbitrary strip of height y and width dx,
then dA (area of the elementary strip) = ydx,
where, y = f (x)
...
1
360
MATHEMATICS
This area is called the elementary area which is located at an arbitrary position
within the region which is specified by some value of x between a and b
...
Symbolically, we express
b
A = ✁ a dA
b
b
✁ a ydx ✁ a f ( x) dx
The area A of the region bounded by
the curve x = g (y), y-axis and the lines y = c,
y = d is given by
d
d
A = ✄ c xdy ✂ ✄ c g ( y) dy
Here, we consider horizontal strips as shown in
the Fig 8
...
2
Remark If the position of the curve under consideration is below the x-axis, then since
f (x) < 0 from x = a to x = b, as shown in Fig 8
...
But, it is only the numerical
value of the area which is taken into consideration
...
e
...
a
Fig 8
...
4
...
Therefore, the area
A bounded by the curve y = f (x), x-axis and the ordinates x = a and x = b is given
by A = | A1 | + A2
...
4
Example 1 Find the area enclosed by the circle x2 + y2 = a2
...
5, the whole area enclosed
by the given circle
= 4 (area of the region AOBA bounded by
the curve, x-axis and the ordinates x = 0 and
x = a) [as the circle is symmetrical about both
x-axis and y-axis]
=4
= 4
a
0
a
0
ydx (taking vertical strips)
a 2 ✁ x 2 dx
Since x2 + y2 = a2 gives
Fig 8
...
Integrating, we get
the whole area enclosed by the given circle
☎x
= 4✟
✡2
a
a2 ✝ x2 ✞
a2
x✆
sin –1 ✠
2
a ☛0
✌✎ a
✍
a2
✜ a2 ✢ ✜ ✣ ✢
☞1 ✏
2
= 4 ✔ ✖ ✑ 0 ✒ sin 1✗ ✓ 0✕ = 4 ✥
✦ ✥ ✦ ✤ ✣a
2
✙
✧ 2 ★ ✧ 2★
✚✘ 2
✛
362
MATHEMATICS
Alternatively, considering horizontal strips as shown in Fig 8
...
6
x2
y2
Example 2 Find the area enclosed by the ellipse 2 ✣ 2 ✤ 1
a
b
Solution From Fig 8
...
So, the required area is
Now
a
= 4✲ 0
b 2
a ✱ x 2 dx
a
a
4b ✳ x 2 2 a 2
–1 x ✴
= ✷ a ✵ x ✶ sin ✸ (Why?)
a ✹2
a ✺0
2
=
4b ✼ ✾ a
a 2 ✻1 ✿ ✽
sin 1❆ ❂ 0 ❄
❃❅ ❀ 0 ❁
a ❉❃❇ 2
2
❈
❊❄
=
4b a 2 ✜
✢ ✜ ab
a 2 2
Fig 8
...
8, the area of the ellipse is
= 4
b
0
b
a
xdy = 4 ✂ b2 ✁ y 2 dy (Why?)
b0
b
b2
y☎
4a ✄ y 2
2
b
y
sin –1 ✟
✆
✝
=
✞
b ✠2
2
b ✡0
☞
✍
4 a ☛✌ b
b2
0
sin –1 1✕ ✑ 0 ✓
✎
✏
✒
✔
= b
2
✗
✘✒✖ 2
✙✓
=
Fig 8
...
2
...
Equations of above mentioned curves will be
in their standard forms only as the cases in other forms go beyond the scope of this
textbook
...
Solution Since the given curve represented by
the equation y = x2 is a parabola symmetrical
about y-axis only, therefore, from Fig 8
...
9
✣ area of the region BONB bounded by curve, y ✢ axis ✤
2✥
✦
★ and the lines y ✧ 0 and y = 4
✩
4
= 2
4
0
3
4
32
2✪ ✫
ydy = 2 ✬ ✭ y 2 ✮ ✱ ✲ 8 ✱
3
3
3 ✯✭ ✰✮
(Why?)
0
Here, we have taken horizontal strips as indicated in the Fig 8
...
364
MATHEMATICS
Alternatively, we may consider the vertical
strips like PQ as shown in the Fig 8
...
To this
end, we solve the equations x2 = y and y = 4
which gives x = –2 and x = 2
...
Therefore, the area of the region AOBA
2
=
✁ 2
Fig 8
...
Henceforth, we shall
consider either of these two, most preferably vertical strips
...
Y
Solution The given equations are
y= x
...
(2)
and
x + y = 32
(4,4)
Solving (1) and (2), we find that the line
and the circle meet at B(4, 4) in the first
quadrant (Fig 8
...
Draw perpendicular
A
X'
X
BM to the x-axis
...
Now, the area of the region OBMO
=
4
✁0
4
ydx ✗ ✁ xdx
0
1 2 4
= ✘✚ x ✙✛ 0 = 8
2
...
11
APPLICATION OF INTEGRALS
365
Again, the area of the region BMAB
=
=
=
4 2
✆
4
ydx =
4 2
✞
1
x 32 ✡ x 2
✌
✎2
✟
✑
✖
✘
32 ✝ x 2 dx
4
☛
1
2
32 ☞ sin
☞
–1
1
1
✒
4 2 ✓ 0 ✔ ✓ 32 ✓ sin –1 1 ✗
✙
2
2
4 2
x
✠
✍
4 2 ✏4
✑
✕
✖
✘
4
32 ✕ 16
2
✔
1
–1 1 ✒
✓ 32 ✓ sin
✗
2
2✙
= 8 ✚ – (8 + 4✚) = 4✚ – 8
Adding (3) and (4), we get, the required area = 4✚
...
(4)
x2 y 2
✛
✜ 1 and the ordinates x = 0
a 2 b2
and x = ae, where, b2 = a2 (1 – e2) and e < 1
...
12) of the region BOB✢RFSB is enclosed by the
ellipse and the lines x = 0 and x = ae
...
12
2
= ab ✷✻✽ e 1 ✹ e
✺
sin –1 e ✸
✼
✾
EXERCISE 8
...
Find the area of the region bounded by the curve y2 = x and the lines x = 1,
x = 4 and the x-axis
...
Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the
first quadrant
...
Find the area of the region bounded by x2 = 4y, y = 2, y = 4 and the y-axis in the
first quadrant
...
Find the area of the region bounded by the ellipse
x2
16
y2
✁1
...
Find the area of the region bounded by the ellipse
x2
4
y2
✁1
...
Find the area of the region in the first quadrant enclosed by x-axis, line x = 3 y
and the circle x2 + y2 = 4
...
2
8
...
7
...
Find the area of the region bounded by the parabola y = x2 and y = x
...
Find the area bounded by the curve x2 = 4y and the line x = 4y – 2
...
Find the area of the region bounded by the curve y2 = 4x and the line x = 3
...
12
...
Area of the region bounded by the curve y = 4x, y-axis and the line y = 3 is
9
4
(C)
9
3
(D)
9
2
8
...
Suppose we are given two curves represented by
y = f (x), y = g (x), where f (x) ☎ g(x) in [a, b] as shown in Fig 8
...
Here the points of
intersection of these two curves are given by x = a and x = b obtained by taking
common values of y from the given equation of two curves
...
As indicated in the Fig 8
...
13
dA = [f (x) – g(x)] dx, and the total area A can be taken as
b
A=
✟a
[f ( x) ✞ g ( x)] dx
Alternatively,
A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]
– [area bounded by y = g (x), x-axis and the lines x = a, x = b]
b
=
✡a
b
b
f ( x) dx ✠ ✡ g ( x) dx = ✟ ☛ f ( x) ✞ g ( x )☞ dx, where f (x) ✌ g (x) in [a, b]
a
a
If f (x) ✌ g (x) in [a, c] and f (x) ✍ g (x) in [c, b], where a < c < b as shown in the
Fig 8
...
14
368
MATHEMATICS
Example 6 Find the area of the region bounded by the two parabolas y = x2 and y2 = x
...
15
...
Therefore, the required area of the shaded region
1
=
f ( x ) ✂ g ( x )✁ dx
✄0
1
=
✆
✡ 0✟
x
☛2
2
✞ x ✝ dx ✌ ✎
✠
✎3
✑
3
x2 ✍
x3 ☞
✏
3 ✒✏
1
Fig 8
...
=
Solution The given equation of the circle x 2 + y 2 = 8x can be expressed as
(x – 4)2 + y2 = 16
...
Its intersection
P (4, 4)
with the parabola y2 = 4x gives
x2 + 4x = 8x
or
x2 – 4x = 0
or
x (x – 4) = 0
X
X
C (4, 0)
Q (8, 0)
O
or
x = 0, x = 4
Thus, the points of intersection of these
two curves are O(0, 0) and P(4,4) above the
x-axis
...
16, the required area of
the region OPQCO included between these
Fig 8
...
17, AOBA is the part of the ellipse 9x2 + y2 = 36 in the first
quadrant such that OA = 2 and OB = 6
...
Solution Given equation of the ellipse 9x2 + y2 = 36 can be expressed as
x2
22
✖
y2
62
✢ 1 and
hence, its shape is as given in Fig 8
...
Accordingly, the equation of the chord AB is
or
6✧0
( x ✧ 2)
0✧2
y = – 3(x – 2)
or
y = – 3x + 6
y–0=
Area of the shaded region as shown in the Fig 8
...
= 3✩
2
0
✪x
= 3✮
✰2
✲2
= 3✷
✹2
2
4 ★ x 2 dx ★ ✩ (6 ★ 3x )dx (Why?)
0
2
2
4 ✬ x2
✴0✵
✭
Fig 8
...
Solution Let A (1, 0), B (2, 2) and C (3, 1) be
the vertices of a triangle ABC (Fig 8
...
Area of ✝ABC
= Area of ✝ ABD + Area of trapezium
BDEC – Area of ✝AEC
Now equation of the sides AB, BC and
CA are given by
y = 2 (x – 1), y = 4 – x, y =
Fig 8
...
2
2
3
3x 1
dx
area of ✝ ABC = ✂ 1 2 ( x 1) dx ✁ ✂ 2 (4 x) dx ✂ 1
2
Hence,
2
3
3
✄ x2 ☎ ✄
☎
x2 ☎
1 ✄ x2
= 2 ✟ ✆ x ✠ ✞ ✟ 4 x ✆ ✠ ✆ ✟ ✆ x✠
2 ☛2 2 ✡ 2
✡2
☛1 ✡
☛1
☞✍ 2 2
✌ ☞✍
✎ 1
☞✍ 2 ✎
✌
32 ✎ ✍
22 ✎ ✌
✏ 2 ✖ ✏ ✍✕ ✏ 1✖✎✔ ✑ ✓✕ 4 ✒ 3 ✏ ✖ ✏ ✕ 4 ✒ 2 ✏ ✖ ✔ – 1 ✓✕ 3 ✏ 3 ✖ ✏ ✕✍ 1 ✏ 1✖✎ ✔
2
2
2
2
✗
✘
✗
✘
✗
✘
✗
✘
✙
✚ ✙
✚ 2 ✙✗ 2
✘ ✗ 2 ✘✚
= 2 ✓✕
3
2
Example 10 Find the area of the region enclosed between the two circles: x2 + y2 = 4
and (x – 2)2 + y2 = 4
...
(1)
x2 + y2 = 4
and
(x – 2)2 + y2 = 4
...
Equation (2) is a circle with
centre C (2, 0) and radius 2
...
19
...
19
APPLICATION OF INTEGRALS
371
Required area of the enclosed region O ACA✂ O between circles
= 2 [area of the region ODCAO]
(Why?)
= 2 [area of the region ODAO + area of the region DCAD]
1
2
= 2 ☎ ✟ 0 y dx ✄ ✟ 1 y dx ✁✆
✝
✞
1
2
2
= 2 ✠✌ ✑ 0 4 ☛ ( x ☛ 2) dx ☞ ✑ 1
4 ☛ x 2 dx ✡
(Why?)
✍
✏
✎
1
✒1
( x ✔ 2) 4 ✔ ( x ✔ 2) 2
= 2✛
✤2
✦1
+ 2✫
✭2
✗
1
–1 ✕ x ✔ 2 ✖ ✓
✘ 4sin ✙
✚✜
2
✢ 2 ✣✥0
2
x 4 ★ x2
✩
1
–1 x ✧
✪ 4sin
2
2 ✬✮1
1
=
✒
✛( x ✔
✤
x ✔ 2 ✖✓
2) 4 ✔ ( x ✔ 2) ✗ 4sin ✙
✚✜
✢ 2 ✣✥ 0
=
✰✲
✸✶✴
✼✺
3 ✵ 4sin –1 ✶
=
✾❁
❈❆❃
●❊
3 ❃ 4❄
=
❏
◆▲
P
=
8❘
❙2 3
3
2
–1 ✕
✗
✒
x
✛
✤
2
4 ✔ x2
✱ ✰
✲ ✴1 ✳ ✳
–1
✯1
✷ ✷ ✴ 4sin ( ✴1) ✹ ✵ ✸ 4sin 1 ✴
✺ 2 ✻✻
✽ ✼
3▲
❀❂
❀✿ ✾
❀
❇ ❅ 4❄ ❉ ❅ ❈ 4❄ ❃
6❋
2❍ ●
2
3 ❃ 4❄
✗ 4sin
–1
3 ✴ 4sin ✯1
x✓
2 ✜✥1
1✱
2 ✹✽
❀✿
6 ❉❍
2■
2■ ❑
❑ ❏
▼ 2■ ❖ ▼ ◆ 2■ ▲ 3 ▲
❖
3
3 ◗
◗ P
EXERCISE 8
...
Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y
...
Find the area bounded by curves (x – 1)2 + y2 = 1 and x2 + y2 = 1
...
Find the area of the region bounded by the curves y = x2 + 2, y = x, x = 0 and
x = 3
...
Using integration find the area of region bounded by the triangle whose vertices
are (– 1, 0), (1, 3) and (3, 2)
...
Using integration find the area of the triangular region whose sides have the
equations y = 2x + 1, y = 3x + 1 and x = 4
...
6
...
Area lying between the curves y2 = 4x and y = 2x is
2
3
(A)
(B)
1
3
(C)
1
4
(D)
3
4
Miscellaneous Examples
Example 11 Find the area of the parabola y2 = 4ax bounded by its latus rectum
...
20, the vertex of the parabola
y2 = 4ax is at origin (0, 0)
...
Also, parabola is
❨
▲
symmetrical about the x-axis
...
20
0
3
=
☞
✌
8
8
a ✍ a 2 ✎ = a2
3
3
✍
✎
✏
✑
Example 12 Find the area of the region bounded
by the line y = 3x + 2, the x-axis and the ordinates
x = –1 and x = 1
...
21, the line
y = 3x + 2 meets x-axis at x =
✔
lies below x-axis for x ✖✗ ✓1,
✙
2 ✕
,1
...
21
❳
APPLICATION OF INTEGRALS
The required area = Area of the region ACBA + Area of the region ADEA
2
3 (3 x ✁
✂ 1
=
2)dx ✁ ✂
1
2
3
(3 x ✁ 2)dx
✄2
2
2
1
☎ 3x
✆ 3
☎ 3x
✆
✝ 2 x✟
✝✞
✝ 2 x✟
✞
✠ 2
✡ ✄1
✠ 2
✡ ✄2
=
=
1 25 13
☛
☞
6 6
3
3
Example 13 Find the area bounded by
the curve y = cos x between x = 0 and
x = 2✌
...
22, the required
area = area of the region OABO + area
of the region BCDB + area of the region
DEFD
...
22
Thus, we have the required area
✍
=
2
✏0
cos x dx ✎
3✍
2
✏ ✍
cos x dx
✎✏
2
✕
2
= ✑ sin x✒ 0
3✕
2
✖ ✓ sin x ✔ ✕
2✕
✖ ✑ sin x ✒
3✕
2
2
2✍
3✍
2
cos x dx
=1+2+1=4
Example 13 Prove that the curves y2 = 4x and x2 = 4y
divide the area of the square bounded by x = 0, x = 4,
y = 4 and y = 0 into three equal parts
...
23
373
374
MATHEMATICS
shown in the Fig 8
...
Now, the area of the region OAQBO bounded by curves y2 = 4x and x2 = 4y
...
(1)
3
3
3
Again, the area of the region OPQAO bounded by the curves x2 = 4y, x = 0, x = 4
and x-axis
=
4
x2
1
16
...
(3)
From (1), (2) and (3), it is concluded that the area of the region OAQBO = area of
the region OPQAO = area of the region OBQRO, i
...
, area bounded by parabolas
y2 = 4x and x2 = 4y divides the area of the square in three equal parts
...
This region is the intersection of the
following regions
...
24
The points of intersection of y = x2 + 1 and y = x + 1 are points P(0, 1) and Q(1, 2)
...
24, the required region is the shaded region OPQRSTO whose area
= area of the region OTQPO + area of the region TSRQT
1
=
✚0
2
( x 2 ✙ 1) dx ✙ ✚ ( x ✙ 1) dx
1
(Why?)
APPLICATION OF INTEGRALS
1
✄✁
=
✂ x3
☎
✆✞
☛✠ 3
=
✎✑ 1
✏ ✎
✒
✌
✗ ✕ ✓ 1✖ ✔ 0 ✘ ✓ ✗ 2 ✓
✚
✛✙ 3
✜ ✛
✂
☎ ✆✞
✡ ☞ 0 ☛✠
x ✟✝
x2
2
☎
✄✁
375
2
x ✟✝
✡ ☞1
✑1
2✍ ✔ ✕
✙2
✒✏
✓ 1✖ ✘
✚✜
=
23
6
Miscellaneous Exercise on Chapter 8
1
...
Find the area between the curves y = x and y = x2
...
Find the area of the region lying in the first quadrant and bounded by y = 4x2,
x = 0, y = 1 and y = 4
...
Sketch the graph of y = x ✢ 3 and evaluate
0
✥✣ 6
x ✤ 3 dx
...
Find the area bounded by the curve y = sin x between x = 0 and x = 2✦
...
Find the area enclosed between the parabola y2 = 4ax and the line y = mx
...
Find the area enclosed by the parabola 4y = 3x2 and the line 2y = 3x + 12
...
Find the area of the smaller region bounded by the ellipse
line
★1
and the
x y
✩ ✪1
...
Find the area of the smaller region bounded by the ellipse
line
x2 y 2
✧
9
4
x2 y 2
✫
✬ 1 and the
a 2 b2
x y
✩ ✪1
...
Find the area of the region enclosed by the parabola x2 = y, the line y = x + 2 and
the x-axis
...
Using the method of integration find the area bounded by the curve x ✢ y ✭ 1
...
376
MATHEMATICS
12
...
13
...
14
...
Find the area of the region {(x, y) : y2 ✆ 4x, 4x2 + 4y2 ✆ 9}
Choose the correct answer in the following Exercises from 16 to 20
...
Area bounded by the curve y = x3, the x-axis and the ordinates x = – 2 and x = 1 is
(A) – 9
15
4
(B)
(C)
15
4
(D)
17
4
17
...
18
...
The area bounded by the y-axis, y = cos x and y = sin x when 0 ✝ x ✝
(A) 2 ( 2 ✞ 1)
(B)
2 ✟1
(C)
2 ✠1
(D)
✄
2
is
2
Summary
✡ The area of the region bounded by the curve y = f (x), x-axis and the lines
b
b
x = a and x = b (b > a) is given by the formula: Area ☛ ☞ ydx ☛ ☞ f ( x) dx
...
c
c
APPLICATION OF INTEGRALS
377
The area of the region enclosed between two curves y = f (x), y = g (x) and
the lines x = a, x = b is given by the formula,
✄ ✆ ✁ f (x) ☎ g (x)✂ dx , where, f (x) ✝ g (x) in [a, b]
If f (x) ✝ g (x) in [a, c] and f (x) ✞ g (x) in [c, b], a < c < b, then
Area ✡ ✌ ✟ f ( x ) ☛ g ( x )✠ dx ☞ ✌ ✟ g ( x) ☛ f ( x) ✠ dx
...
This method arose in the solution of problems
on calculating areas of plane figures, surface areas and volumes of solid bodies
etc
...
The greatest development of method of exhaustion in the early
period was obtained in the works of Eudoxus (440 B
...
) and Archimedes
(300 B
...
)
Systematic approach to the theory of Calculus began in the 17th century
...
Newton introduced the basic notion of inverse function
called the anti derivative (indefinite integral) or the inverse method of tangents
...
In 1696, he followed a suggestion made by J
...
This corresponded to Newton’s inverse method of tangents
...
However, respective theories accomplished results that
were practically identical
...
Conclusively, the fundamental concepts and theory of Integral Calculus
and primarily its relationships with Differential Calculus were developed in the
work of P
...
Newton and G
...
✍
378
MATHEMATICS
However, this justification by the concept of limit was only developed in the
works of A
...
Cauchy in the early 19th century
...
The discovery
that differentiation and integration are inverse operations belongs to Newton
and Leibnitz”
...
– C
...
PEIRCE
13
...
We discussed the axiomatic approach formulated by
Russian Mathematician, A
...
Kolmogorov (1903-1987)
and treated probability as a function of outcomes of the
experiment
...
On the basis of this
relationship, we obtained probabilities of events associated
with discrete sample spaces
...
In this chapter, we shall discuss
the important concept of conditional probability of an event
given that another event has occurred, which will be helpful
in understanding the Bayes' theorem, multiplication rule of
Pierre de Fermat
probability and independence of events
...
In the last
section of the chapter, we shall study an important discrete probability distribution
called Binomial distribution
...
13
...
If we have two events from the same sample space, does the information
about the occurrence of one of the events affect the probability of the other event? Let
us try to answer this question by taking up a random experiment in which the outcomes
are equally likely to occur
...
The sample space of the
experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
532
MATHEMATICS
1
to each sample point
...
Then
E = {HHH, HHT, HTH, THH}
and
F = {THH, THT, TTH, TTT}
Therefore
P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})
Since the coins are fair, we can assign the probability
1 1 1 1 1
✁ (Why ?)
8 8 8 8 2
P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})
=
and
1 1 1 1 1
✂ ✂ ✂ ✄
8 8 8 8 2
E ☎ F = {THH}
=
Also
1
8
Now, suppose we are given that the first coin shows tail, i
...
F occurs, then what is
the probability of occurrence of E? With the information of occurrence of F, we are
sure that the cases in which first coin does not result into a tail should not be considered
while finding the probability of E
...
In other words, the additional information really
amounts to telling us that the situation may be considered as being that of a new
random experiment for which the sample space consists of all those outcomes only
which are favourable to the occurrence of the event F
...
with
P(E ☎ F) = P({THH}) =
Thus, Probability of E considering F as the sample space =
1
,
4
1
4
This probability of the event E is called the conditional probability of E given
that F has already occurred, and is denoted by P (E|F)
...
e
...
Thus
P(E|F) =
PROBABILITY
533
Thus, we can also write the conditional probability of E given that F has occurred as
P(E|F) =
=
Number of elementaryeventsfavourable to E F
Number of elementaryevents which arefavourable to F
n (E ✁ F)
n (F)
Dividing the numerator and the denominator by total number of elementary events
of the sample space, we see that P(E|F) can also be written as
n(E ✂ F)
n(S)
P(E|F) =
n(F)
n(S)
✄ P(E ✂ F)
P(F)
...
e
...
e
...
2
...
Example 1 If P (A) =
7
9
4
, P (B) =
and P (A ✞ B) =
, evaluate P (A|B)
...
What is the probability that both the children are
boys given that at least one of them is a boy ?
PROBABILITY
535
Solution Let b stand for boy and g for girl
...
If it is known that the number on the drawn card is
more than 3, what is the probability that it is an even number?
Solution Let A be the event ‘the number on the card drawn is even’ and B be the
event ‘the number on the card drawn is greater than 3’
...
Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Then
A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10}
and
A ✂ B = {4, 6, 8, 10}
Also
Then
P(A) =
5
7
4
, P ( B) = and P (A ✄ B) ☎
10
10
10
4
P (A ✆ B) 10
✝7
P(A|B) =
P ( B)
10
✝4
7
Example 4 In a school, there are 1000 students, out of which 430 are girls
...
What is the probability that a student
chosen randomly studies in Class XII given that the chosen student is a girl?
Solution Let E denote the event that a student chosen randomly studies in Class XII
and F be the event that the randomly chosen student is a girl
...
536
MATHEMATICS
Now
P(F) =
Then
P(E|F) =
430
1000
0
...
043 (Why?)
1000
P (E ✂ F) 0
...
1
P ( F)
0
...
Events A and B are defined as below:
A : 4 on the third throw
B : 6 on the first and 5 on the second throw
Find the probability of A given that B has already occurred
...
Now
and
Now
Then
☎(1,1,4) (1,2,4)
...
(2,6,4) ✆
✝
✝
A = ✞(3,1,4) (3,2,4)
...
(4,6,4) ✟
✝(5,1,4) (5,2,4)
...
(6,6,4) ✝
✠
✡
B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}
A ☛ B = {(6,5,4)}
...
What is the conditional probability that the number 4 has appeared at least
once?
Solution Let E be the event that ‘number 4 appears at least once’ and F be the event
that ‘the sum of the numbers appearing is 6’
...
However, the same definition can also be used in
the general case where the elementary events of the sample space are not equally
likely, the probabilities P (E ✂ F) and P (F) being calculated accordingly
...
Example 7 Consider the experiment of tossing a coin
...
Find the
conditional probability of the event that ‘the die shows
a number greater than 4’ given that ‘there is at least
one tail’
...
The sample space of the experiment may be
described as
Fig 13
...
Thus, the probabilities assigned to the 8 elementary
events
(H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6)
1 1 1 1 1 1 1 1
, , , , , , ,
respectively which is
4 4 12 12 12 12 12 12
clear from the Fig 13
...
are
Fig 13
...
Then
Now
F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}
E = {(T,5), (T,6)} and E ✂ F = {(T,5), (T,6)}
P (F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)})
+ P ({(T,4)}) + P({(T,5)}) + P({(T,6)})
=
and
Hence
1 1 1 1 1 1 1 3
✁
4 12 12 12 12 12 12 4
P (E ✂ F) = P ({(T,5)}) + P ({(T,6)}) =
1 1 1
✁
12 12 6
1
P (E ✄ F) 6 2
☎ ☎
P(E|F) =
3 9
P (F)
4
EXERCISE 13
...
Given that E and F are events such that P(E) = 0
...
3 and
P(E ✂ F) = 0
...
Compute P(A|B), if P(B) = 0
...
32
3
...
8, P (B) = 0
...
4, find
(ii) P(A|B)
(iii) P(A ✆ B)
(i) P (A ✂ B)
4
...
6
...
If P(A) =
PROBABILITY
7
...
539
Two coins are tossed once, where
E : tail appears on one coin,
F : one coin shows head
E : no tail appears,
F : no head appears
A die is thrown three times,
E : 4 appears on the third toss,
F : 6 and 5 appears respectively
on first two tosses
9
...
A black and a red dice are rolled
...
(b) Find the conditional probability of obtaining the sum 8, given that the red die
resulted in a number less than 4
...
A fair die is rolled
...
Assume that each born child is equally likely to be a boy or a girl
...
An instructor has a question bank consisting of 300 easy True / False questions,
200 difficult True / False questions, 500 easy multiple choice questions and 400
difficult multiple choice questions
...
Given that the two numbers appearing on throwing two dice are different
...
15
...
Find the conditional probability
of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’
...
If P (A) =
1
, P (B) = 0, then P (A|B) is
2
(A) 0
(C) not defined
1
2
(D) 1
(B)
540
MATHEMATICS
17
...
3 Multiplication Theorem on Probability
Let E and F be two events associated with a sample space S
...
In other words, E ✂ F denotes the
simultaneous occurrence of the events E and F
...
Very often we need to find the probability of the event EF
...
The probability of event EF is obtained
by using the conditional probability as obtained below :
We know that the conditional probability of event E given that F has occurred is
denoted by P(E|F) and is given by
P(E|F) =
P (E F)
,P (F) ✁ 0
P (F)
From this result, we can write
P (E ✂ F) = P (F)
...
(1)
Thus,
P(E ✂ F) = P(E)
...
(2)
Combining (1) and (2), we find that
P (E ✂ F) = P(E) P(F|E)
= P(F) P(E|F) provided P(E) ✄ 0 and P(F) ✄ 0
...
Let us now take up an example
...
Two balls are drawn from the
urn one after the other without replacement
...
We have to find P (E ✂ F) or P (EF)
...
e
...
Therefore, the probability that the
second ball drawn is black, given that the ball in the first draw is black, is nothing but
the conditional probability of F given that E has occurred
...
e
...
The following example illustrates the extension of multiplication rule of probability
for three events
...
What is the probability that first two cards are kings and the
third card drawn is an ace?
Solution Let K denote the event that the card drawn is king and A be the event that
the card drawn is an ace
...
Now there are three kings in (52 ❾ 1) = 51 cards
...
Now there are four aces in left 50 cards
...
4 Independent Events
Consider the experiment of drawing a card from a deck of 52 playing cards, in which
the elementary events are assumed to be equally likely
...
We also have
Since P(E) =
1
P (E ✄ F) 52 1
☎
☎ ☎ P(F)
P(F|E) =
1 13
P(E)
4
1
= P (F|E) shows that occurrence of event E has not affected
13
the probability of occurrence of the event F
...
Such events are called independent events
...
P (F|E)
...
(2)
P (E ✂ F) = P (E)
...
P (F)
Remarks
(i) Two events E and F are said to be dependent if they are not independent, i
...
if
P (E ✂ F ) ✄ P (E)
...
Term ‘independent’ is defined in terms of ‘probability of events’
whereas mutually exclusive is defined in term of events (subset of sample space)
...
Clearly, ‘independent’ and
‘mutually exclusive’ do not have the same meaning
...
e
...
(iii) Two experiments are said to be independent if for every pair of events E and F,
where E is associated with the first experiment and F with the second experiment,
the probability of the simultaneous occurrence of the events E and F when the
two experiments are performed is the product of P(E) and P(F) calculated
separately on the basis of two experiments, i
...
, P (E ✂ F) = P (E)
...
Example 10 A die is thrown
...
P (F)
E and F are independent events
...
Let the event A be ‘odd number on the
first throw’ and B the event ‘odd number on the second throw’
...
Solution If all the 36 elementary events of the experiment are considered to be equally
likely, we have
18 1
18 1
and P (B)
36 2
36 2
P (A ✂ B) = P (odd number on both throws)
P(A) =
Also
=
Now
Clearly
Thus,
P (A) P(B) =
9 1
✄
36 4
1 1 1
☎ ✄
2 2 4
P (A ✂ B) = P (A) × P (B)
A and B are independent events
Example 12 Three coins are tossed simultaneously
...
Of the pairs (E,F),
(E,G) and (F,G), which are independent? which are dependent?
Solution The sample space of the experiment is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Clearly
E = {HHH, TTT}, F= {HHH, HHT, HTH, THH}
PROBABILITY
and
Also
Therefore
and
G = {HHT, HTH, THH, HTT, THT, TTH, TTT}
E ✂ F = {HHH}, E ✂ G = {TTT}, F ✂ G = { HHT, HTH, THH}
P (E) =
P (E ✂ F) =
2
8
1
, P (F)
4
1
, P (E ✁ G)
8
4
8
1
, P(G)
2
7
8
1
, P(F ✁ G)
8
3
8
Also
P (E)
...
P(G) =
1 7
✄
2 8
7
16
Thus
545
1 7
✄
4 8
7
32
P (E ✂ F) = P(E)
...
P(G)
and
P (F ✂ G) ✆ P (F)
...
Example 13 Prove that if E and F are independent events, then so are the events
E and F✞✝
Solution Since E and F are independent, we have
P (E ✂ F) = P (E)
...
3, it is clear
that E ✂ F and E ✂ F ✞ are mutually exclusive events
and also E =(E ✂ F) ✟ (E ✂ F ✞)
...
(1)
✡
✭✡ ’ ❋ ’ ✮
❋
❙
P (E) = P (E ✂ F) + P(E ✂ F ✞)
P (E ✂ F ✞) = P(E) ❾ P(E ✂ F)
= P(E) ❾ P(E)
...
P(F ✞)
Hence, E and F ✞ are independent
✭✡ ❋ ’ ✮
✭✡ ❋✮
Fig 13
...
P(A )
= 1 P(A ) + P(B) P(A )
= 1 P(A ) [1 P(B)]
= 1 P(A ) P (B )
1
...
2
1
, find P (A
5
B) if A and B are independent events
...
Two cards are drawn at random and without replacement from a pack of 52
playing cards
...
3
...
If all the three oranges are good, the box is approved
for sale, otherwise, it is rejected
...
4
...
Let A be the event ‘head appears on
the coin’ and B be the event ‘3 on the die’
...
5
...
Let A be the event,
‘the number is even,’ and B be the event, ‘the number is red’
...
Let E and F be events with P (E)
E and F independent?
✄
3
, P (F)
5
✄
3
and P (E
10
✂
F) =
1
...
Find p if they are (i) mutually exclusive (ii) independent
...
Let A and B be independent events with P (A) = 0
...
4
...
Given that the events A and B are such that P(A) =
9
...
1
7
1
, P(B) =
and P(not A or not B) =
...
Given two independent events A and B such that P(A) = 0
...
6
...
A die is tossed thrice
...
13
...
Find the probability that
(i) both balls are red
...
(iii) one of them is black and other is red
...
Events A and B are such that P (A) =
1
1
and
2
3
respectively
...
15
...
In which of
the following cases are the events E and F independent ?
(i) E : ‘the card drawn is a spade’
F : ‘the card drawn is an ace’
(ii) E : ‘the card drawn is black’
F : ‘the card drawn is a king’
(iii) E : ‘the card drawn is a king or queen’
F : ‘the card drawn is a queen or jack’
...
Probability of solving specific problem independently by A and B are
548
MATHEMATICS
16
...
A student is selected
at random
...
(b) If she reads Hindi news paper, find the probability that she reads English
news paper
...
Choose the correct answer in Exercises 17 and 18
...
The probability of obtaining an even prime number on each die, when a pair of
dice is rolled is
1
1
1
(A) 0
(B)
(C)
(D)
3
12
36
18
...
5 Bayes' Theorem
Consider that there are two bags I and II
...
One ball is drawn at random from one of the
1
bags
...
e
...
In
other words, we can find the probability that the ball drawn is of a particular colour, if
we are given the bag from which the ball is drawn
...
Famous mathematician, John Bayes' solved the problem
of finding reverse probability by using conditional probability
...
Before stating and proving the Bayes' theorem, let us first take up a definition and
some preliminary results
...
5
...
, En is said to represent a partition of the sample space S if
(a) Ei ✂ Ej = ☎, i ✄ j, i, j = 1, 2, 3,
...
✆ En= S and
(c) P(Ei) > 0 for all i = 1, 2,
...
In other words, the events E1, E2,
...
As an example, we see that any nonempty event E and its complement E✞ form a
partition of the sample space S since they satisfy E ✂ E✞ = ☎ and E ✆ E✞ = S
...
3, one can easily observe that if E and F are any
two events associated with a sample space S, then the set {E ✂ F✞, E ✂ F, E✞ ✂ F, E✞ ✂ F✞}
is a partition of the sample space S
...
There can be several partitions of the same sample space
...
13
...
2 Theorem of total probability
Let {E1, E2,
...
, En has nonzero probability of occurrence
...
+ P(En) P(A|En)
n
=
✄ P(E j ) P (A|E j )
j ✁1
Proof Given that E1, E2,
...
4)
...
(1)
S = E1 ✆ E2 ✆ ✝✝✝ ✆ En
and
Ei ✂ Ej = ☎, i ✟ j, i, j = 1, 2,
...
✆ En)
Fig 13
...
We know that
Ei and Ej are disjoint, for i ✠ j , therefore, A ✂ Ei and A ✂ Ej are also disjoint for all
i ✟ j, i, j = 1, 2,
...
Thus,
P(A) = P [(A ✂ E1) ✆ (A ✂ E2)✆
...
+ P (A ✂ En)
Now, by multiplication rule of probability, we have
P(A ✂ Ei) = P(Ei) P(A|Ei) as P (Ei) ✟ 0✡i = 1,2,
...
+ P (En)P(A|En)
or
P(A) =
n
✁ P(E j ) P (A|E j )
j 1
Example 15 A person has undertaken a construction job
...
65
that there will be strike, 0
...
32 that the construction job will be completed on time if there is a
strike
...
Solution Let A be the event that the construction job will be completed on time, and B
be the event that there will be a strike
...
We have
P(B) = 0
...
65 = 0
...
32, P(A|B✞) = 0
...
65 × 0
...
35 × 0
...
208 + 0
...
488
Thus, the probability that the construction job will be completed in time is 0
...
We shall now state and prove the Bayes' theorem
...
, En are n non empty events which constitute a partition
of sample space S, i
...
E1, E2 ,
...
✆ En = S and
A is any event of nonzero probability, then
P(Ei|A) =
P (E i ) P (A|E i )
n
for any i = 1, 2, 3,
...
The events E1, E2,
...
The probability P(Ei) is called the priori probability of the hypothesis Ei
The conditional probability P(Ei |A) is called a posteriori probability of the
hypothesis Ei
...
Since the
Ei's are a partition of the sample space S, one and only one of the events Ei occurs (i
...
one of the events Ei must occur and only one can occur)
...
e
...
The Bayes' theorem has its applications in variety of situations, few of which are
illustrated in following examples
...
One ball is drawn at random from one of the bags and it is found to
be red
...
Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II
and A be the event of drawing a red ball
...
In
box I, both coins are gold coins, in box II, both are silver coins and in the box III, there
is one gold and one silver coin
...
If the coin is of gold, what is the probability that the other coin in the box is also of gold?
552
MATHEMATICS
Solution Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively
...
= P(E 1|A)
By Bayes' theorem, we know that
P(A|E3) = P(a gold coin from bag III) =
P(E1|A) =
P(E1 ) P(A|E1 )
P(E1 ) P(A|E1 ) + P(E 2 ) P(A|E 2 ) + P(E 3 ) P (A|E 3 )
1
1
2
3
✁
=
1
1
1 1 3
1✂ 0 ✂
3
3
3 2
Example 18 Suppose that the reliability of a HIV test is specified as follows:
Of people having HIV, 90% of the test detect the disease but 10% go undetected
...
From a large population of which only 0
...
What is the probability that the person actually has HIV?
Solution Let E denote the event that the person selected is actually having HIV and A
the event that the person's HIV test is diagnosed as +ive
...
Also E✞ denotes the event that the person selected is actually not having HIV
...
We are given that
P(E) = 0
...
1
✄ 0
...
999
P(A|E) = P(Person tested as HIV+ive given that he/she
is actually having HIV)
90
0
...
01
100
Now, by Bayes' theorem
P(E|A) =
P(E) P(A|E)
P(E) P(A|E) + P(E ✁) P (A|E ✁)
0
...
9
90
✄
0
...
9 ☎ 0
...
01 1089
= 0
...
=
Thus, the probability that a person selected at random is actually having HIV
given that he/she is tested HIV+ive is 0
...
Example 19 In a factory which manufactures bolts, machines A, B and C manufacture
respectively 25%, 35% and 40% of the bolts
...
A bolt is drawn at random from the product and is found
to be defective
...
Let the event E be ‘the bolt is defective’
...
Given that,
P(B1) = 25% = 0
...
35 and P(B3) = 0
...
05
Similarly, P(E|B2) = 0
...
02
...
35 0
...
25 0
...
35 0
...
40 0
...
0140 28
✂
0
...
From the past experience, it is known that
the probabilities that he will come by train, bus, scooter or by other means of transport
3 1 1
2
1 1
1
are respectively , , and
...
When he arrives, he is late
...
3
1
1
2
, P (T2 ) ✂ , P (T3 ) ✂ and P (T4 ) ✂
Then
P(T1) =
(given)
10
5
10
5
1
P(E|T1) = Probability that the doctor arriving late comes by train =
4
1
1
Similarly, P(E|T2) = , P(E|T3) =
and P(E|T4) = 0, since he is not late if he
3
12
comes by other means of transport
...
2
PROBABILITY
555
Example 21 A man is known to speak truth 3 out of 4 times
...
Find the probability that it is actually a six
...
Then
P(S1) = Probability that six occurs =
1
6
5
6
P(E|S1) = Probability that the man reports that six occurs when six has
actually occurred on the die
P(S2) = Probability that six does not occur =
3
4
P(E|S2) = Probability that the man reports that six occurs when six has
not actually occurred on the die
= Probability that the man speaks the truth =
= Probability that the man does not speak the truth 1 ✁
3
4
1
4
Thus, by Bayes' theorem, we get
P(S1|E) = Probability that the report of the man that six has occurred is
actually a six
=
P(S1 ) P(E |S1 )
P(S1 ) P(E|S1 ) + P(S2 ) P (E|S2 )
1 3
✂
1 24 3
6 4
✄ ✂ ✄
=
1 3 5 1 8 8 8
✂ ☎ ✂
6 4 6 4
3
Hence, the required probability is
...
3
1
...
A ball is drawn at random, its colour is
noted and is returned to the urn
...
What is the probability that
the second ball is red?
556
2
...
4
...
6
...
8
...
MATHEMATICS
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black
balls
...
Find the probability that the ball is drawn from the
first bag
...
Previous year results report that 30% of all
students who reside in hostel attain A grade and 20% of day scholars attain A
grade in their annual examination
...
Let
be the probability that he knows the answer and
4
4
be the probability that he guesses
...
What is the probability that the stu4
dent knows the answer given that he answered it correctly?
A laboratory blood test is 99% effective in detecting a certain disease when it is
in fact, present
...
5% of
the healthy person tested (i
...
if a healthy person is tested, then, with probability
0
...
If 0
...
One is a two headed coin (having head on both faces),
another is a biased coin that comes up heads 75% of the time and third is an
unbiased coin
...
The probability of an accidents are 0
...
03 and 0
...
One of the insured persons meets with an accident
...
Past record shows that machine A produced
60% of the items of output and machine B produced 40% of the items
...
All the items are put into one stockpile and then one item is chosen at
random from this and is found to be defective
...
The probabilities that the first and the second groups will win are
PROBABILITY
557
0
...
4 respectively
...
7 and the corresponding probability is 0
...
Find the probability that the new product introduced was by
the second group
...
Suppose a girl throws a die
...
If she gets 1, 2, 3 or 4, she tosses a coin once and
notes whether a head or tail is obtained
...
A manufacturer has three machine operators A, B and C
...
A is on the job for 50% of the
time, B is on the job for 30% of the time and C is on the job for 20% of the time
...
A card from a pack of 52 cards is lost
...
Find the probability of
the lost card being a diamond
...
A coin is tossed
...
The probability that actually there was head is
13
...
If A and B are two events such that A ✟ B and P(B) ✄ 0, then which of the
following is correct?
(A)
(A) P(A | B)
P (B)
P (A)
(C) P(A|B) ☛ P(A)
(B) P(A⑤B) < P(A)
(D) None of these
13
...
In
most of these experiments, we were not only interested in the particular outcome that
occurs but rather in some number associated with that outcomes as shown in following
examples/experiments
...
(ii) In tossing a coin 50 times, we may want the number of heads obtained
...
In all of the above experiments, we have a rule which assigns to each outcome of
the experiment a single real number
...
Hence, it is a variable
...
A random
variable is usually denoted by X
...
A random variable can take any real value, therefore, its
co-domain is the set of real numbers
...
For example, let us consider the experiment of tossing a coin two times in succession
...
If X denotes the number of heads obtained, then X is a random variable and for
each outcome, its value is as given below :
X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0
...
For
example, let Y denote the number of heads minus the number of tails for each outcome
of the above sample space S
...
Thus, X and Y are two different random variables defined on the same sample
space S
...
For each head, he is
given Rs 2 by the organiser of the game and for each tail, he has to give Rs 1
...
Let X denote the amount gained or lost by the person
...
Solution X is a number whose values are defined on the outcomes of a random
experiment
...
Now, sample space of the experiment is
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
PROBABILITY
Then
559
X (HHH) = Rs (2 × 3) = Rs 6
X (HHT) = X (HTH) = X (THH) = Rs (2 × 2 ❾ 1 × 1
...
50
X (HTT) = X (THT) = (TTH) = Rs (1 × 2) – (2 × 1
...
50) =
Rs 4
...
Thus, for each element of the sample
space, X takes a unique value, hence, X is a function on the sample space whose range
is
{– 1, 2
...
50, 6}
Example 23 A bag contains 2 white and 1 red balls
...
The process is repeated again
...
Solution Let the balls in the bag be denoted by w1, w2, r
...
13
...
1 Probability distribution of a random variable
Let us look at the experiment of selecting one family out of ten families f1, f2 ,
...
Let the families f1, f2,
...
Let us select a family and note down the number of members in the family denoting
X
...
Now, X will take the value 2 when the family f4 is selected
...
Similarly,
X = 4, when family f2, f6 or f9 is selected,
X = 5, when family f5 or f10 is selected
and
X = 6, when family f8 is selected
...
10
1
1
...
and
P(X = 6) = P({f8}) =
In general, the probability distribution of a random variable X is defined as follows:
Definition 5 The probability distribution of a random variable X is the system of numbers
X
:
x1
x2
...
pn
n
where,
pi ✁ 0,
✂
pi = 1, i = 1, 2,
...
, xn are the possible values of the random variable X and
pi (i = 1,2,
...
e
...
Hence, the probability that
X takes value xi is always nonzero, i
...
P(X = xi) 0
...
Hence, the sum of all the probabilities in a probability distribution
must be one
...
Find the probability distribution of the number of aces
...
Let it be denoted by X
...
Now, since the draws are done with replacement, therefore, the two draws form
independent experiments
...
P(non-ace) + P (non-ace)
...
562
MATHEMATICS
Solution Let X denote the number of doublets
...
6
36
Probability of getting a doublet
Probability of not getting a doublet
Now
1
6
1✁
1
6
5
6
5 5 5 125
✂ ✂ ✄
6 6 6 216
P(X = 1) = P (one doublet and two non-doublets)
P(X = 0) = P (no doublet) =
=
1 5 5 5 1 5 5 5 1
☎ ☎ ✆ ☎ ☎ ✆ ☎ ☎
6 6 6 6 6 6 6 6 6
✝ 1 52 ✞ 75
= 3✡ ✟ 2 ☛ ✠
☞ 6 6 ✌ 216
P(X = 2) = P (two doublets and one non-doublet)
1 1 5 1 5 1 5 1 1
✍ 1 5 ✎ 15
✏ ✏ ✑ ✏ ✏ ✑ ✏ ✏ ✒ 3✓ 2 ✏ ✔ ✒
✕ 6 6 ✖ 216
6 6 6 6 6 6 6 6 6
P(X = 3) = P (three doublets)
=
and
1 1 1
1
✂ ✂ ✄
6 6 6 216
Thus, the required probability distribution is
=
X
0
1
2
3
P(X)
125
216
75
216
15
216
1
216
Verification Sum of the probabilities
n
125
75
15
1
✙
✙
✙
✘ pi =
216 216 216 216
i ✗1
=
125 ✆ 75 ✆ 15 ✆ 1
216
216
1
216
PROBABILITY
563
Example 26 Let X denote the number of hours you study during a randomly selected
school day
...
✁0
...
(b) What is the probability that you study at least two hours ? Exactly two hours? At
most two hours?
Solution The probability distribution of X is
X
0
1
2
3
4
P(X)
0
...
1 + k + 2k + 2k + k = 1
i
...
k = 0
...
15 = 0
...
15 = 0
...
1 + k + 2k = 0
...
1 + 3 × 0
...
55
13
...
2 Mean of a random variable
In many problems, it is desirable to describe some feature of the random variable by
means of a single number that can be computed from its probability distribution
...
In this section, we shall discuss mean only
...
564
MATHEMATICS
Definition 6 Let X be a random variable whose possible values x1, x2, x3,
...
, pn , respectively
...
e
...
The mean of a random variable X is also called the expectation of X, denoted by
E(X)
...
+ xn pn
...
Example 27 Let a pair of dice be thrown and the random variable X be the sum of the
numbers that appear on the two dice
...
Solution The sample space of the experiment consists of 36 elementary events in the
form of ordered pairs (xi , yi), where xi = 1, 2, 3, 4, 5, 6 and yi = 1, 2, 3, 4, 5, 6
...
e
...
Now
P(X = 2) = P({(1,1)})
☎
1
36
P(X = 3) = P({(1,2), (2,1)})
✆
2
36
P(X = 4) = P({(1,3), (2,2), (3,1)})
☎
3
36
P(X = 5) = P({(1,4), (2,3), (3,2), (4,1)}) ☎
4
36
P(X = 6) = P({(1,5), (2,4), (3,3), (4,2), (5,1)}) ✆
5
36
P(X = 7) = P({(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}) ☎
P(X = 8) = P({(2,6), (3,5), (4,4), (5,3), (6,2)}) ☎
5
36
6
36
PROBABILITY
P(X = 9) = P({(3,6), (4,5), (5,4), (6,3)})
P(X = 10) = P({(4,6), (5,5), (6,4)})
565
4
36
3
36
2
36
P(X = 11) = P({(5,6), (6,5)})
1
36
The probability distribution of X is
P(X = 12) = P({(6,6)})
X or xi
2
3
4
5
6
7
8
9
10
11
12
P(X) or pi
1
36
2
36
3
36
4
36
5
36
6
36
5
36
4
36
3
36
2
36
1
36
Therefore,
n
1
2
3
4
➭ = E(X) = ✆ xi pi ✂ 2✄ ☎ 3✄ ☎ 4✄ ☎ 5✄
36
36
36
36
i ✁1
✝6✞
=
5
6
5
4
3
2
1
✝7✞ ✝8✞
✝ 9 ✞ ✝ 10 ✞ ✝ 11✞ ✝ 12 ✞
36
36
36
36
36
36
36
2 ✟ 6 ✟ 12 ✟ 20 ✟ 30 ✟ 42 ✟ 40 ✟ 36 ✟ 30 ✟ 22 ✟ 12
=7
36
Thus, the mean of the sum of the numbers that appear on throwing two fair dice is 7
...
6
...
In fact, if the variance is small, then the values of the
random variable are close to the mean
...
X
1
2
3
4
P(X)
1
8
2
8
3
8
2
8
566
MATHEMATICS
Y
–1
0
4
5
6
P(Y)
1
8
2
8
3
8
1
8
1
8
1
2
3
2 22
✁ 2 ✁ 3 ✁ 4 ✂ ✂ 2
...
75
8
8
8
8
8 8
The variables X and Y are different, however their means are same
...
5)
...
5
To distinguish X from Y, we require a measure of the extent to which the values of
the random variables spread out
...
Likewise, the variability or spread in the
values of a random variable may be measured by variance
...
,xn occur with
probabilities p(x1), p(x2),
...
Let ➭ = E (X) be the mean of X
...
Another formula to find the variance of a random variable
...
Solution The sample space of the experiment is S = {1, 2, 3, 4, 5, 6}
...
Then X is a random variable
which can take values 1, 2, 3, 4, 5, or 6
...
Find the mean, variance and standard deviation
of the number of kings
...
X is a random
variable which can assume the values 0, 1 or 2
...
37
221
EXERCISE 13
...
State which of the following are not the probability distributions of a random
variable
...
(i)
X
0
P(X) 0
...
1
1
2
0
...
2
1
2
0
...
2 – 0
...
3
570
MATHEMATICS
(iii)
Y
–1
P(Y)
(iv)
Z
P(Z)
0
...
1
0
...
3
0
...
4
0
...
05
2
...
Two balls are randomly drawn
...
What are the possible values of X? Is X a
random variable ?
3
...
What are possible values of X?
4
...
(ii) number of tails in the simultaneous tosses of three coins
...
5
...
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn
at random with replacement
...
7
...
If the coin is
tossed twice, find the probability distribution of number of tails
...
A random variable X has the following probability distribution:
X
0
P(X) 0
Determine
(i) k
(iii) P(X > 6)
1
k
7
2 3 4 5 6
2
2
2k 2k 3k k 2k 7k 2 +k
(ii) P(X < 3)
(iv) P(0 < X < 3)
PROBABILITY
571
9
...
(b) Find P (X < 2), P (X ✍ 2), P(X ☛ 2)
...
Find the mean number of heads in three tosses of a fair coin
...
Two dice are thrown simultaneously
...
12
...
Let X denote the larger of the two numbers obtained
...
13
...
Find the variance and standard deviation of X
...
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20,
17, 16, 19 and 20 years
...
What is the probability distribution of the random variable X? Find
mean, variance and standard deviation of X
...
In a meeting, 70% of the members favour and 30% oppose a certain proposal
...
Find E(X) and Var (X)
...
The mean of the numbers obtained on throwing a die having written 1 on three
faces, 2 on two faces and 5 on one face is
8
3
17
...
Let X be the
number of aces obtained
...
7 Bernoulli Trials and Binomial Distribution
13
...
1 Bernoulli trials
Many experiments are dichotomous in nature
...
In such cases, it is customary to call one of the outcomes a ‘success’
and the other ‘not success’ or ‘failure’
...
Each time we toss a coin or roll a die or perform any other experiment, we call it a
trial
...
The outcome of any trial is independent of the
outcome of any other trial
...
Such independent trials which have only two outcomes usually
referred as ‘success’ or ‘failure’ are called Bernoulli trials
...
(ii) The trials should be independent
...
(iv) The probability of success remains the same in each trial
...
Obviously, the successive throws
of the die are independent experiments
...
written on six faces, then p =
2
2
Example 30 Six balls are drawn successively from an urn containing 7 red and 9 black
balls
...
(i) replaced
Solution
(i) The number of trials is finite
...
Hence, the drawing of balls with replacements are Bernoulli trials
...
e
...
Clearly, the probability of success is
15
not same for all trials, hence the trials are not Bernoulli trials
...
7
...
Let S and F denote respectively success and failure in each
trial
...
Clearly, six different cases are there as listed below:
SFFFFF, FSFFFF, FFSFFF, FFFSFF, FFFFSF, FFFFFS
...
It will be
lengthy job to list all of these ways
...
,
n number of successes may be lengthy and time consuming
...
For this purpose, let us take the
experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for
success and failure respectively in each trial
...
The probability distribution of the number of successes is as below :
P(X = 0) = P(no success)
= P({FFF}) = P(F) P(F) P(F)
= q
...
q = q3 since the trials are independent
P(X = 1) = P(one successes)
= P({SFF, FSF, FFS})
= P({SFF}) + P({FSF}) + P({FFS})
= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)
= p
...
q + q
...
q + q
...
p = 3pq2
P(X = 2) = P (two successes)
= P({SSF, SFS, FSS})
= P({SSF}) + P ({SFS}) + P({FSS})
574
MATHEMATICS
= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)
= p
...
q
...
q
...
p
...
P(S)
...
Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1
...
, n successes can be obtained as 1st, 2nd,
...
To prove this assertion (result), let us find the probability of x-successes in
an experiment of n-Bernoulli trials
...
n!
Now, x successes (S) and (n – x) failures (F) can be obtained in x !(n ✁ x)! ways
...
P(n–x) failures is
P (S)
...
P(S) ✄ P (F)
...
P(F)
= ☎✆✆✝✆✆✞ ☎✆✆✝✆✆✞ = px qn–x
x times
( n ✂ x ) times
Thus, the probability of x successes in n-Bernoulli trials is
or nCx px qn–x
Thus
P(x successes) = n C x p x q n ✂ x ,
n!
px qn–x
x !(n ✁ x)!
x = 0, 1, 2,
...
(q = 1 – p)
Clearly, P(x successes), i
...
n C x p x q n ✂ x is the (x + 1)th term in the binomial
expansion of (q + p)n
...
Hence, this
PROBABILITY
575
distribution of number of successes X can be written as
X
0
1
2
...
C x q n–xp x
n
n
Cn pn
The above probability distribution is known as binomial distribution with parameters
n and p, because for given values of n and p, we can find the complete probability
distribution
...
, n
...
A binomial distribution with n-Bernoulli trials and probability of success in each
trial as p, is denoted by B (n, p)
...
Example 31 If a fair coin is tossed 10 times, find the probability of
(i) exactly six heads
(ii) at least six heads
(iii) at most six heads
Solution The repeated tosses of a coin are Bernoulli trials
...
1
Clearly, X has the binomial distribution with n = 10 and p =
2
Therefore
P(X = x) = nCxqn–xpx, x = 0, 1, 2,
...
Find the probability that there is at least one defective egg
...
Since the
drawing is done with replacement, the trials are Bernoulli trials
...
100 10
9
10
P(at least one defective egg) = P(X ✙ 1) = 1 – P (X = 0)
q = 1✗ p ✘
10
9✁
☎
10
✆
✝
10
= 1 ✚ C0 ✄
= 1✛
910
1010
EXERCISE 13
...
A die is thrown 6 times
...
A pair of dice is thrown 4 times
...
3
...
What is the probability
that a sample of 10 items will include not more than one defective item?
4
...
What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is a spade?
5
...
05
...
6
...
If four balls
are drawn successively with replacement from the bag, what is the probability
that none is marked with the digit 0?
7
...
Suppose a student
tosses a fair coin to determine his answer to each question
...
Find the probability
that he answers at least 12 questions correctly
...
Suppose X has a binomial distribution B ✂ 6,
☎
1✁
✄
...
(Hint : P(X = 3) is the maximum among all P(xi), xi = 0,1,2,3,4,5,6)
9
...
A person buys a lottery ticket in 50 lotteries, in each of which his chance of
1
...
Find the probability of getting 5 exactly twice in 7 throws of a die
...
Find the probability of throwing at most 2 sixes in 6 throws of a single die
...
It is known that 10% of certain articles manufactured are defective
...
In a box containing 100 bulbs, 10 are defective
...
The probability that a student is not a swimmer is
(D)
9
10
1
...
The colour of the ball is black, what is the probability that ball drawn is from the
box III?
PROBABILITY
579
Solution Let A, E1, E2, E3 and E4 be the events as defined below :
A : a black ball is selected
E1 : box I is selected
E3 : box III is selected
E2 : box II is selected
E4 : box IV is selected
Since the boxes are chosen at random,
Therefore
P(E1) = P(E2) = P(E3) = P(E4) =
1
4
3
2
1
4
, P(A|E2) = , P(A|E3) =
and P(A|E4) =
18
8
7
13
P(box III is selected, given that the drawn ball is black) = P(E3|A)
...
165
✆
Example 34 Find the mean of the Binomial distribution B ✞ 4,
✠
1✝
✟
...
✘ 3✙
✔
4
P(X = x) = C x ✖
We know that
i
...
the distribution of X is
xi
4
0
1
xi P(xi)
P(x i)
4
2✛
✢
✣ 3✤
4
✚
C0 ✜
2✕
✗
✘ 3✙
✔
C1 ✖
3
1✕
✖ ✗
✘ 3✙
✔
0
4
2✕
✗
✘ 3✙
✔
C1 ✖
3
✔
✖
✘
1✕
✗
3✙
1☞
✍
...
How many minimum
4
number of times must he/she fire so that the probability of hitting the target at least
once is more than 0
...
Obviously, n fires are n Bernoulli trials
...
Then P(X = x) = C x q p ★ C x ✂ ✄
4
☎ 4✆
Now, given that,
P(hitting the target at least once) > 0
...
e
...
99
n✧ x
3✁
✂ ✄
☎ 4✆
x
★
n
Cx
3x
...
99
1
or
n
or
C0
n
C0
1
> 0
...
01 i
...
n < 0
...
01
The minimum value of n to satisfy the inequality (1) is 4
...
4n >
or
581
...
Find their respective probabilities of winning, if A starts first
...
Thus,
P(S) =
1
5
, P(F) ✂
6
6
1
6
A gets the third throw, when the first throw by A and second throw by B result into
failures
...
✠ 6✡ ✠ 6✡
2
Hence,
4
1 ☎ 5✆ ☎ 1 ✆ ☎ 5✆ ☎ 1✆
P(A wins) = ☞ ✞ ✟ ✞ ✟ ☞ ✞ ✟ ✞ ✟ ☞
...
+ arn–1 +
...
P
...
(Refer A
...
3 of Class XI Text book)
...
If it is
incorrectly set up, it produces only 40% acceptable items
...
If after a certain set up, the machine produces
2 acceptable items, find the probability that the machine is correctly setup
...
Also let B1 represent the event of correct set up and B2 represent the event of
incorrect setup
...
8, P(B2) = 0
...
9 × 0
...
4 × 0
...
8 × 0
...
9
648
✄
✄ 0
...
8 × 0
...
9 + 0
...
4 × 0
...
A and B are two events such that P (A) ☎ 0
...
A couple has two children,
(i) Find the probability that both children are males, if it is known that at least
one of the children is male
...
3
...
25% of women have grey hair
...
What is the probability of this person being male?
Assume that there are equal number of males and females
...
Suppose that 90% of people are right-handed
...
An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15
bear a mark 'Y'
...
If 6 balls are drawn in this way, find the probability that
(i) all will bear 'X' mark
...
(iii) at least one ball will bear 'Y' mark
...
6
...
The probability that he will
clear each hurdle is
5
...
A die is thrown again and again until three sixes are obtained
...
8
...
An experiment succeeds twice as often as it fails
...
10
...
In a game, a man wins a rupee for a six and loses a rupee for any other number
when a fair die is thrown
...
Find the expected value of the amount he wins / loses
...
Suppose we have four boxes A,B,C and D containing coloured marbles as given
below:
Box
Marble colour
Red
White
Black
A
1
6
3
B
6
2
2
C
8
1
1
D
0
6
4
One of the boxes has been selected at random and a single marble is drawn from
it
...
Assume that the chances of a patient having a heart attack is 40%
...
At a time a
patient can choose any one of the two options with equal probabilities
...
Find the probability that the patient followed a course of
meditation and yoga?
14
...
2
15
...
From previous testing procedures, the following probabilities are assumed to be known:
P(A fails) = 0
...
15
P(A and B fail) = 0
...
Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls
...
The ball so drawn is found to be red in colour
...
Choose the correct answer in each of the following:
17
...
If P(A|B) > P(A), then which of the following is correct :
✂ B) < P(A)
...
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
(A) P(B|A) = 1
(B) P(A|B) = 1
(C) P(B|A) = 0
(D) P(A|B) = 0
PROBABILITY
585
Summary
The salient features of the chapter are –
The conditional probability of an event E, given the occurrence of the event F
✂ P (EP(F)✁ F) , P(F) ✄ 0
0 ✍ P (E|F) ✍ 1,
P (E✞|F) = 1 – P (E|F)
P ((E ✆ F)|G) = P (E|G) + P (F|G) – P ((E ☎ F)|G)
P (E ☎ F) = P (E) P (F|E), P (E) ✄ 0
P (E ☎ F) = P (F) P (E|F), P (F) ✄ 0
is given by P (E | F)
If E and F are independent, then
P (E F) = P (E) P (F)
P (E|F) = P (E), P (F) 0
P (F|E) = P (F), P(E) 0
Theorem of total probability
Let {E1, E2,
...
, En has nonzero probability
...
+ P (En) P(A|En)
Bayes' theorem If E1, E2,
...
e
...
, En are pairwise disjoint and E1 E2
...
The probability distribution of a random variable X is the system of numbers
x2
...
pn
☞ ✎ p ✌ 1, i ✌ 1, 2,
...
, xn occur with
probabilities p1, p2, p3,
...
The mean of X, denoted by , is
➭
✂✁ x p
...
Let X be a random variable whose possible values x1, x2,
...
, p(xn) respectively
...
The variance of X, denoted by Var (X) or
➭
✎ , is defined as ✝ ☎ Var (X) = ✂✁ ( x ✆ ✄ )
or equivalently ✎ = E (X – ➭)
n
2
2
x
x
2
i
p ( xi )
i 1
2
2
x
The non-negative number
☞✠
x
☛✟ ( x ✡ ✞)
n
Va r (X) =
i
2
p( xi )
i 1
is called the standard deviation of the random variable X
...
(ii) The trials should be independent
...
(iv) The probability of success remains the same in each trial
...
, n
(q = 1 – p)
Historical Note
The earliest indication on measurement of chances in game of dice appeared
in 1477 in a commentary on Dante's Divine Comedy
...
In this treatise, he gives the number of favourable cases
for each event when two dice are thrown
...
Galileo analysed that when three dice are
thrown, the sum of the number that appear is more likely to be 10 than the sum 9,
because the number of cases favourable to 10 are more than the number of
cases for the appearance of number 9
...
A French gambler, Chevalier de Metre asked Pascal to explain
some seeming contradiction between his theoretical reasoning and the
observation gathered from gambling
...
Pascal solved
the problem in algebraic manner while Fermat used the method of combinations
...
The next great work on probability theory is by Jacob Bernoulli (1654-1705),
in the form of a great book, "Ars Conjectendi" published posthumously in 1713
by his nephew, Nicholes Bernoulli
...
The next
remarkable work on probability lies in 1993
...
N
...
His book, ‘Foundations of
probability’ published in 1933, introduces probability as a set function and is
considered a ‘classic!’
...
– D
...
1 Introduction
In Class XI and in Chapter 5 of the present book, we
discussed how to differentiate a given function f with respect
to an independent variable, i
...
, how to find f (x) for a given
function f at each x in its domain of definition
...
(1)
An equation of the form (1) is known as a differential
equation
...
Henri Poincare
(1854-1912 )
These equations arise in a variety of applications, may it be in Physics, Chemistry,
Biology, Anthropology, Geology, Economics etc
...
In this chapter, we will study some basic concepts related to differential equation,
general and particular solutions of a differential equation, formation of differential
equations, some methods to solve a first order - first degree differential equation and
some applications of differential equations in different areas
...
2 Basic Concepts
We are already familiar with the equations of the type:
x2 – 3x + 3 = 0
sin x + cos x = 0
x+y=7
...
(2)
...
(4)
dx
We see that equations (1), (2) and (3) involve independent and/or dependent variable
(variables) only but equation (4) involves variables as well as derivative of the dependent
variable y with respect to the independent variable x
...
x
In general, an equation involving derivative (derivatives) of the dependent variable
with respect to independent variable (variables) is called a differential equation
...
g
...
(5)
✄ ☎ ✆ = 0 is an ordinary differential equation
dx 2 ✝ dx ✞
Of course, there are differential equations involving derivatives with respect to
more than one independent variables, called partial differential equations but at this
stage we shall confine ourselves to the study of ordinary differential equations only
...
2
✟ Note
1
...
For derivatives of higher order, it will be inconvenient to use so many dashes
as supersuffix therefore, we use the notation yn for nth order derivative
dny
...
2
...
Order of a differential equation
Order of a differential equation is defined as the order of the highest order derivative of
the dependent variable with respect to the independent variable involved in the given
differential equation
...
(6)
DIFFERENTIAL EQUATIONS
d2y
dx 2
y =0
381
...
(8)
The equations (6), (7) and (8) involve the highest derivative of first, second and
third order respectively
...
9
...
2 Degree of a differential equation
To study the degree of a differential equation, the key point is that the differential
equation must be a polynomial equation in derivatives, i
...
, y✟, y✠, y✠✟ etc
...
(9)
2
✒ dy ✓ ✒ dy ✓
2
✖ dx ✗ ✔ ✖ dx ✗ ✕ sin y = 0
✘ ✙ ✘ ✙
...
(11)
We observe that equation (9) is a polynomial equation in y✠✟✦ y✠ and y✟, equation (10)
is a polynomial equation in y✟ (not a polynomial in y though)
...
But equation (11) is not a polynomial equation in y✟ and
degree of such a differential equation can not be defined
...
In view of the above definition, one may observe that differential equations (6), (7),
(8) and (9) each are of degree one, equation (10) is of degree two while the degree of
differential equation (11) is not defined
...
382
MATHEMATICS
Example 1 Find the order and degree, if defined, of each of the following differential
equations:
(i)
(iii)
dy
dx
2
d2y
dy
✂ dy ✄
✝0
(ii) xy 2 ☎ x ✞ ✟ ✆ y
dx
✠ dx ✡
dx
cos x ✁ 0
y☞☞☞ ✌ y 2 ✌ e y☛ ✍ 0
Solution
(i) The highest order derivative present in the differential equation is
dy
, so its
dx
order is one
...
(ii) The highest order derivative present in the given differential equation is
its order is two
...
power raised to
dx 2
(iii) The highest order derivative present in the differential equation is y✏✏✏ , so its
order is three
...
EXERCISE 9
...
1
...
y✎ + 5y = 0
d 2s
✔ ds ✕
3
s
✖
✗0
✙
✚ dt ✛
dt 2
3
...
✣ 2 ✤ ✌ cos ✣ ✤ ✍ 0
✥ dx ✦
✥ dx ✦
5
...
( y☞☞☞) 2 + (y✧)3 + (y✎)4 + y5 = 0
7
...
y✂ + y = ex
9
...
y✄ + 2y✂ + sin y = 0
11
...
The order of the differential equation
d2y
dy
✡3
☛ y ☞ 0 is
2
dx
dx
(A) 2
(B) 1
(D) not defined
2x 2
(C) 0
(D) not defined
9
...
General and Particular Solutions of a Differential Equation
In earlier Classes, we have solved the equations of the type:
...
(2)
sin x – cos x = 0
Solution of equations (1) and (2) are numbers, real or complex, that will satisfy the
given equation i
...
, when that number is substituted for the unknown x in the given
equation, L
...
S
...
H
...
d2y
✌ y✍0
...
e
...
H
...
becomes equal to R
...
S
...
Consider the function given by
...
When this function and its derivative are substituted in equation (3),
L
...
S
...
H
...
So it is a solution of the differential equation (3)
...
(5)
When this function and its derivative are substituted in equation (3) again
L
...
S
...
H
...
Therefore ✎1 is also a solution of equation (3)
...
Whereas function ☎1 contains no
arbitrary constants but only the particular values of the parameters a and b and hence
is called a particular solution of the given differential equation
...
The solution free from arbitrary constants i
...
, the solution obtained from the general
solution by giving particular values to the arbitrary constants is called a particular
solution of the differential equation
...
Differentiating both sides of equation with respect
to x , we get
dy
✆ ✝3 e✄3 x
dx
Now, differentiating (1) with respect to x, we have
...
H
...
= 9 e– 3x + (–3e– 3x) – 6
...
H
...
Therefore, the given function is a solution of the given differential equation
...
(1)
Differentiating both sides of equation (1) with respect to x, successively, we get
dy
= – a sin x + b cos x
dx
d2y
= – a cos x – b sin x
dx 2
DIFFERENTIAL EQUATIONS
385
d2y
and y in the given differential equation, we get
dx 2
L
...
S
...
H
...
Therefore, the given function is a solution of the given differential equation
...
2
In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a
solution of the corresponding differential equation:
: y✄ – y ✂ = 0
1
...
y = x + 2x + C
3
...
y = Ax
:
xy
1 ✁ x2
xy✂ = y (x ✝ 0)
6
...
xy = log y + C
:
8
...
x + y = tan–1y
:
:
4
...
The number of arbitrary constants in the general solution of a differential equation
of fourth order are:
(B) 2
(C) 3
(D) 4
(A) 0
12
...
y =
a 2 ✞ x 2 x ✟ (–a, a) :
x+y
9
...
...
(2)
which is a differential equation
...
5
...
)]
that this equation represents the family of circles and one member of the family is the
circle given in equation (1)
...
(3)
x2 + y2 = r 2
By giving different values to r, we get different members of the family e
...
x2 + y2 = 1, x2 + y2 = 4, x2 + y2 = 9 etc
...
1)
...
We are interested in finding a differential equation
that is satisfied by each member of the family
...
This
equation is obtained by differentiating equation (3) with
respect to x, i
...
,
2x + 2y
dy
=0
dx
or
x+y
dy
=0
dx
...
1
which represents the family of concentric circles given by equation (3)
...
(5)
By giving different values to the parameters m and c, we get different members of
the family, e
...
,
y=x
(m = 1, c = 0)
y=
3x
(m =
3 , c = 0)
y=x+1
(m = 1, c = 1)
y=–x
(m = – 1, c = 0)
y=–x–1
(m = – 1, c = – 1) etc
...
2)
...
We are now interested in finding a differential equation that is satisfied by each
member of the family
...
This is obtained by differentiating equation (5) with
respect to x, successively we get
dy
dx
d2y
✞ m , and
dx 2
✟0
✂
✂
...
Note that equations (3) and (5) are the general
solutions of equations (4) and (6) respectively
...
2
9
...
1 Procedure to form a differential equation that will represent a given
family of curves
(a) If the given family F1 of curves depends on only one parameter then it is
represented by an equation of the form
F1 (x, y, a) = 0
...
Differentiating equation (1) with respect to x, we get an equation involving
y✠, y, x, and a, i
...
,
g (x, y, y✠, a) = 0
...
(3)
F (x, y, y✠) = 0
(b) If the given family F2 of curves depends on the parameters a, b (say) then it is
represented by an equation of the from
F2 (x, y, a, b) = 0
...
e
...
(5)
But it is not possible to eliminate two parameters a and b from the two equations
and so, we need a third equation
...
(6)
388
MATHEMATICS
The required differential equation is then obtained by eliminating a and b from
equations (4), (5) and (6) as
...
Example 4 Form the differential equation representing the family of curves y = mx,
where, m is arbitrary constant
...
(1)
Differentiating both sides of equation (1) with respect to x, we get
dy
=m
dx
dy
x
Substituting the value of m in equation (1) we get y
dx
dy
x
or
–y=0
dx
which is free from the parameter m and hence this is the required differential equation
...
Solution We have
y = a sin (x + b)
...
(2)
...
(4)
dx 2
which is free from the arbitrary constants a and b and hence this the required differential
equation
...
Solution We know that the equation of said family
of ellipses (see Fig 9
...
(1)
Fig 9
...
(2)
Differentiating both sides of equation (2) with respect to x, we get
✡ dy
☛
x ☞y
✍ dy
✡ y ☛ ✡ d 2 y ☛ ✌ dx
=0
✌✏ ✍✑ ✌✏ 2 ✍✑ ✎ ✌✏
✍
x
dx
x 2 ✑ dx
2
or
xy
d2y
dy
✒ dy ✓
✔x✕ ✖ – y
=0
✗ dx ✘
dx
dx 2
...
❨
Example 7 Form the differential equation of the family
of circles touching the x-axis at origin
...
Let (0, a) be the coordinates of the
centre of any member of the family (see Fig 9
...
Therefore, equation of family C is
x2 + (y – a)2 = a2 or x2 + y2 = 2ay
...
Differentiating both
sides of equation (1) with respect to x,we get
2x ✙ 2 y
dy
dy
= 2a
dx
dx
❳’
❖
❨’
Fig 9
...
(2)
dy
dx
Substituting the value of a from equation (2) in equation (1), we get
x2 + y2 =
✂ x ☎ y dy ✄
✆
dx ✝✟
2y ✞
dy
dx
dy 2
dy
( x ✠ y 2 ) = 2 xy ✠ 2 y 2
dx
dx
or
dy
2 xy
= 2 2
dx
x –y
or
This is the required differential equation of the given family of circles
...
Solution Let P denote the family of above said parabolas (see Fig 9
...
Therefore, equation
of family P is
...
(2)
dy ☛
✌ ( x)
dx ✎
dy
=0
dx
which is the differential equation of the given family
of parabolas
...
5
DIFFERENTIAL EQUATIONS
391
EXERCISE 9
...
1
...
6
...
8
...
10
...
x y
✁1
2
...
y = a e3x + b e– 2x
a b
y = e2x (a + bx)
5
...
Form the differential equation of the family of parabolas having vertex at origin
and axis along positive y-axis
...
Form the differential equation of the family of hyperbolas having foci on x-axis
and centre at origin
...
Which of the following differential equations has y = c1 ex + c2 e–x as the general
solution?
d2y
d2y
d2y
d2y
✂ y ✄ 0 (B)
☎ y ✄ 0 (C)
✂ 1 ✄ 0 (D)
☎1 ✄ 0
2
2
2
dx
dx 2
dx
dx
12
...
5
...
9
...
1 Differential equations with variables separable
A first order-first degree differential equation is of the form
dy
= F (x, y)
dx
...
The differential equation (1) then has the form
dy
= h (y)
...
(2)
dx
If h (y) ✝ 0, separating the variables, (2) can be rewritten as
1
dy = g (x) dx
h( y )
...
(4)
Thus, (4) provides the solutions of given differential equation in the form
H (y) = G (x) + C
1
Here, H (y) and G (x) are the anti derivatives of h ( y ) and g (x) respectively and
C is the arbitrary constant
...
(1)
Separating the variables in equation (1), we get
(2 – y) dy = (x + 1) dx
Integrating both sides of equation (2), we get
✞ (2 ✆ y ) dy = ✞ ( x ✟ 1) dx
or
or
or
2y ✠
y2
x2
✡ x ✡ C1
=
2
2
x2 + y2 + 2x – 4y + 2 C1 = 0
x2 + y2 + 2x – 4y + C = 0, where C = 2C1
which is the general solution of equation (1)
...
(2)
DIFFERENTIAL EQUATIONS
Example 10 Find the general solution of the differential equation
393
dy 1 y 2
...
(1)
dx
dy
✄ 1 y 2 = ☎ 1 ✂ x2
or
tan–1 y = tan–1x + C
which is the general solution of equation (1)
...
Solution If y ✝ 0, the given differential equation can be written as
dy
= – 4x dx
y2
Integrating both sides of equation (1), we get
...
2
...
Example 12 Find the equation of the curve passing through the point (1, 1) whose
differential equation is x dy = (2x2 + 1) dx (x ✝ 0)
...
(1)
Integrating both sides of equation (1), we get
✍ dy
=
✞
✎ ✡ 2x ✠
☞
1✟
☛ dx
x✌
y = x2 + log | x | + C
...
Therefore substituting x = 1, y = 1 in equation (2), we
get C = 0
...
or
Example 13 Find the equation of a curve passing through the point (–2, 3), given that
the slope of the tangent to the curve at any point (x, y) is
2x
...
(1)
...
dx
...
By
treating dx and dy like separate entities, we can give neater expressions to many calculations
...
DIFFERENTIAL EQUATIONS
395
Substituting x = –2, y = 3 in equation (3), we get C = 5
...
In
how many years Rs 1000 double itself?
Solution Let P be the principal at any time t
...
(1)
...
(3)
Now
P = 1000, when t = 0
Substituting the values of P and t in (3), we get C = 1000
...
Then
t
2000 = 1000 e 20
✍ t = 20 loge2
EXERCISE 9
...
dy 1 ✎ cos x
✏
dx 1 ✑ cos x
2
...
MATHEMATICS
dy
dx
y ✁ 1 ( y ✂ 1)
4
...
(ex + e–x) dy – (ex – e–x) dx = 0
6
...
y log y dx – x dy = 0
8
...
ex tan y dx + (1 – ex) sec2 y dy = 0
dx
For each of the differential equations in Exercises 11 to 14, find a particular solution
satisfying the given condition:
9
...
( x ☎ x ☎ x ☎ 1)
12
...
cos ✡
14
...
Find the equation of a curve passing through the point (0, 0) and whose differential
equation is y✎ = ex sin x
...
For the differential equation xy
dy
✄ ( x ☎ 2) ( y ☎ 2) , find the solution curve
dx
passing through the point (1, –1)
...
Find the equation of a curve passing through the point (0, –2) given that at any
point (x, y) on the curve, the product of the slope of its tangent and y coordinate
of the point is equal to the x coordinate of the point
...
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the
line segment joining the point of contact to the point (– 4, –3)
...
19
...
If
initially its radius is 3 units and after 3 seconds it is 6 units
...
DIFFERENTIAL EQUATIONS
397
20
...
Find the
value of r if Rs 100 double itself in 10 years (loge2 = 0
...
21
...
An amount
of Rs 1000 is deposited with this bank, how much will it worth after 10 years
(e0 5 = 1
...
22
...
The number is increased by 10% in 2
hours
...
The general solution of the differential equation
dy
dx
✁e
x y
(A) ex + e–y = C
(B) ex + ey = C
(C) e–x + ey = C
(D) e–x + e–y = C
is
9
...
2 Homogeneous differential equations
Consider the following functions in x and y
F1 (x, y) = y2 + 2xy,
✂
F3 (x, y) = cos ☎
✝
y✄
✆,
x✞
F2 (x, y) = 2x – 3y,
F4 (x, y) = sin x + cos y
If we replace x and y by ✟x and ✟y respectively in the above functions, for any nonzero
constant ✟, we get
F1 (✟x, ✟y) =
✟2
(y2 + 2xy) =
✟2
F1 (x, y)
F2 (✟x, ✟y) = ✟ (2x – 3y) = ✟ F2 (x, y)
✂ ✠y ✄
✂
✆ ✡ cos ☎
✝ ✠x ✞
✝
F3 (✟x, ✟y) = cos ☎
y✄
0
✆ = ✟ F3 (x, y)
x✞
F4 (✟x, ✟y) = sin ✟x + cos ✟y ☛
✟n
F4 (x, y), for any n ☞ N
Here, we observe that the functions F1, F 2, F 3 can be written in the form
F(✟x, ✟y) = ✟n F (x, y) but F4 can not be written in this form
...
We note that in the above examples, F1, F2, F3 are homogeneous functions of
degree 2, 1, 0 respectively but F4 is not a homogeneous function
...
To solve a homogeneous differential equation of the type
A differential equation of the form
dy
✥ y✦
✰ F ✮ x, y ✯ = g ★ ✩
dx
✪x✫
We make the substitution
y = v
...
(1)
...
(3)
DIFFERENTIAL EQUATIONS
v
x
dv
= g (v)
dx
dv
= g (v) – v
dx
Separating the variables in equation (4), we get
x
or
399
...
(5)
Integrating both sides of equation (5), we get
1
dv
✂ g (v ) ✁ v = ☎ x dx ✄ C
...
x
dx
✆Note If the homogeneous differential equation is in the form dy ✝ F( x , y )
where, F (x, y) is homogenous function of degree zero, then we make substitution
x
✞ v i
...
, x = vy and we proceed further to find the general solution as discussed
y
dx
x
above by writing
✡ F( x, y ) ✡ h ☛✟ ☞✠
...
Solution The given differential equation can be expressed as
x ✎ 2y
dy
=
x✁ y
dx
Let
Now
F (x, y) =
F (✏ x, ✏ y) =
x✎ 2y
x✁ y
✑ ( x ✒ 2 y) 0
✞ ✑ ✓ f ( x, y )
✑ ( x ✔ y)
...
So, the given differential
equation is a homogenous differential equation
...
(2)
✎
R
...
S
...
Therefore, equation (1) is a homogeneous differential equation
...
(3)
Differentiating equation (3) with respect to, x we get
dy
dv
= v✕x
dx
dx
Substituting the value of y and
dy
in equation (1) we get
dx
v✕ x
dv 1 ✕ 2v
=
1✖ v
dx
or
x
dv 1 ✕ 2v
✖v
=
1✖ v
dx
or
x
dv
v2 ✗ v ✗ 1
=
dx
1✘ v
v ✙1
✛ dx
dv =
x
v ✚ v ✚1
Integrating both sides of equation (5), we get
or
2
v ✙1
dv =
v ✚ v ✚1
✜ 2
or
✖✢
dx
x
1 2v ✚ 1 ✙ 3
dv = – log | x | + C1
2 ✜ v2 ✚ v ✚ 1
...
tan ✍1 ✔
✓ ✒ log x ✎ C1
✖ 3 ✕✗
2
2 3
or
1
1
✏ 2v ✎ 1✑
log v 2 ✎ v ✎ 1 ✎ log x 2 ✓ 3 tan ✍1 ✔
✎C
✖ 3 ✕✗ 1
2
2
Replacing v by
y
, we get
x
or
y2 y
1
1
✚ 2 y ✙ x✛
✙C
log 2 ✙ ✙ 1 ✙ log x 2 ✜ 3 tan ✘1 ✢
✤ 3x ✣✥ 1
x
2
2
x
or
✧ y2 y ★
1
✧ 2y ✩ x ★
log ✫ 2 ✩ ✩ 1✬ x 2 ✪ 3 tan ✦1 ✫
✬ ✩ C1
2
x ✮
✭ 3x ✮
✭x
or
log ( y 2 ✰ xy ✰ x 2 ) ✳ 2 3 tan ✯1 ✴
or
log ( x 2 ✰ xy ✰ y 2 ) ✳ 2 3 tan ✯1 ✴
1✟ ✞ 3 ✟
✠☛ v ☎ 2 ☞✡ ☎ ✠ 2 ✡
☛ ☞
2
2
401
dv ✝ ✆ log x ☎ C1
(Why?)
✱ 2 y ✰ x ✲ 2C
✵✰ 1
✶ 3x ✷
✱ x ✰ 2y ✲
✵ ✰C
✶ 3x ✷
which is the general solution of the differential equation (1)
✸ y ✹ dy
✸ y✹
Example 16 Show that the differential equation x cos ✼ ✽ ✺ y cos ✼ ✽ ✻ x is
✾ x ✿ dx
✾ x✿
homogeneous and solve it
...
(1)
402
MATHEMATICS
It is a differential equation of the form
Here
dy
F( x, y )
...
Therefore, the given differential equation is a homogeneous differential equation
...
(2)
Differentiating equation (2) with respect to x, we get
dv
dy
= v✒ x
dx
dx
Substituting the value of y and
dy
in equation (1), we get
dx
v✒x
dv
v cos v ✒ 1
=
dx
cos v
or
x
dv
v cos v ✒ 1
✓v
=
cos v
dx
or
x
dv
1
=
cos v
dx
or
Therefore
cos v dv =
dx
x
1
✔ cos v dv = ✕ x dx
...
sin ✂
x
✝
x
✞
Example 17 Show that the differential equation 2 y e y dx ✡ ✟✌ y ☛ 2 x e y ✠✍ dy ☞ 0 is
homogeneous and find its particular solution, given that, x = 0 when y = 1
...
(1)
x
y
x
Let
F(x, y) =
2 xe y
✎y
x
2 ye y
Then
F (✏x, ✏ y) =
x
✑
y
✕ ✓ 2 xe ✖
✓
✗
✒
y✔
x
✒
y
✑
✕ ✓ 2 ye ✔
✓
✔
✗
✘
✔
✘
✙✕
0
[F( x, y )]
Thus, F(x, y) is a homogeneous function of degree zero
...
To solve it, we make the substitution
x = vy
...
(3)
Substituting x = 0 and y = 1 in equation (3), we get
2 e0 + log | 1| = C ✞ C = 2
Substituting the value of C in equation (3), we get
x
y
2 e + log | y | = 2
which is the particular solution of the given differential equation
...
2 xy
Solution We know that the slope of the tangent at any point on a curve is
Therefore,
x2 ✝ y2
dy
=
2 xy
dx
dy
...
(1)
Clearly, (1) is a homogenous differential equation
...
5
In each of the Exercises 1 to 10, show that the given differential equation is homogeneous
and solve each of them
...
(x2 + xy) dy = (x2 + y2) dx
3
...
x2
dy
✂ x2 ✄ 2 y 2
dx
xy
2
...
(x2 – y2) dx + 2xy dy = 0
6
...
☞ x cos ✍ ✎ ✠ y sin ✍ ✎✌ y dx ✡ ☞ y sin ✍ ✎ ☛ x cos ✍ ✎✌ x dy
✏ x✑
✏ x ✑✓
✏ x✑
✏ x ✑✓
✒
✒
8
...
✔ y✕
y dx ✗ x log ✙ ✚ dy ✖ 2 x dy ✘ 0
✛ x✜
x
x
✢
✣
x✣
y
y ✢
10
...
(x + y) dy + (x – y) dx = 0; y = 1 when x = 1
12
...
✵ x sin ✳✷ ✴✸ ✰ y ✶ dx ✱ x dy ✲ 0; y ✲ when x = 1
x
4
✹
✺
14
...
2 xy
y2 ✄ 2x2
dy
✂ 0 ; y = 2 when x = 1
dx
16
...
(A) y = vx
(B) v = yx
(C) x = vy
dx
✻x✼
✽ h ✾ ✿ can be solved by
dy
❀ y❁
(D) x = v
DIFFERENTIAL EQUATIONS
407
17
...
5
...
Some examples of the first order linear differential equation are
dy
✁ y = sin x
dx
dy ✂ 1 ✄
☎ ✆ ✝ y = ex
dx ✞ x ✟
1
dy ✠ y ✡
☛☞
=
✌
dx ✍ x log x ✎
x
Another form of first order linear differential equation is
dx
✏ P1 x = Q 1
dy
where, P1 and Q1 are constants or functions of y only
...
(g (x)) y = Q
...
(1)
...
H
...
becomes a derivative of y
...
i
...
g (x)
dy
d
+ P
...
g (x)]
dx
dx
or
g (x)
dy
dy
+ P
...
g (x) = g✂ (x)
✞
g ( x)
g ( x)
Integrating both sides with respect to x, we get
or
P=
✁
or
✁
Pdx =
✄
g ( x)
dx
g ( x)
P ☎ dx = log (g (x))
g (x) = e✆ P dx
or
On multiplying the equation (1) by g(x) = e ✝
P dx
, the L
...
S
...
This function g(x) = e ✝
(I
...
) of the given differential equation
...
✖
dx ✚ C
DIFFERENTIAL EQUATIONS
409
Steps involved to solve first order linear differential equation:
(i) Write the given differential equation in the form
dy
dx
Py ✁ Q where P, Q are
constants or functions of x only
...
F) = e ✂ P dx
...
F) = ✝ ✄ Q × I
...
Then I
...
(I
...
F ✡ dy ☛ C
Example 19 Find the general solution of the differential equation
dy
✌ y ✍ cos x
...
F = e
Multiplying both sides of equation by I
...
(1)
410
MATHEMATICS
=
✁ cos x e
=
✝ cos x
=
✁ cos x e
x
e✄ x
x
x
✁ ✂ sin x e
✝ ☎ sin x (– e
✞
✡ sin x
e
x
dx
✄x
✁
) ✝ ✠ cos x (✝e ✄ x ) dx ✆
✂
✟
cos x e
x
dx
I = – e–x cos x + sin x e–x – I
2I = (sin x – cos x) e–x
or
or
(sin x ☞ cos x ) e☛ x
2
Substituting the value of I in equation (1), we get
or
I=
ye– x =
or
y=
✌ sin x ☞ cos x ✍ ☛ x
✎
✏
✑e
2
✒
✓
✌
✏
✒
C
sin x ☞ cos x ✍
x
✑ ✎ Ce
2
✓
which is the general solution of the given differential equation
...
dx
Solution The given differential equation is
dy
✔ 2 y = x2
dx
Dividing both sides of equation (1) by x, we get
x
...
x
dx
2
log f ( x )
log x
2
✙ f ( x )]
✘ x [ as e
I
...
x2 =
✛
( x ) ( x 2 ) dx ✚ C =
✛
x3 dx ✚ C
x2
✜2
✢Cx
4
which is the general solution of the given differential equation
...
Solution The given differential equation can be written as
dx
dy
x
= 2y
y
This is a linear differential equation of the type
✝ ✆ y dy
1
Q1 = 2y
...
F ✞ e
1
dx
✁ P1 x ✂ Q1 , where P1 ✂ ✄ and
y
dy
☎ 1
✞ e✆ log y ✞ elog ( y ) ✞
1
y
Hence, the solution of the given differential equation is
x
✟1✠
1
= ✎ (2 y ) ☛ ☞ dy ✡ C
y
✌ y✍
or
x
= ✑ (2dy ) ✏ C
y
or
x
= 2y + C
y
or
x = 2y2 + Cy
which is a general solution of the given differential equation
...
Solution The given equation is a linear differential equation of the type
where P = cot x and Q = 2x + x2 cot x
...
F = e ✗
cot x dx
✘ e log sin x ✘ sin x
Hence, the solution of the differential equation is given by
y
...
(1)
in equation (1), we get
2
✏ ✎✑
✓
✔
✕
✖
2
✗✘
✏✎✑
✔✒
✕
✖
sin ✓
2
C
2
4
Substituting the value of C in equation (1), we get
2
y sin x = x sin x ✗
or
2
y= x
✙
✎
✘
2
4
2
4 sin x
(sin x ✚ 0)
which is the particular solution of the given differential equation
...
If the slope
of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate
(abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point
...
(1)
This is a linear differential equation of the type
Therefore,
dy
...
F = e
✣ ✢ x dx
dy
✜ Py ✍ Q , where P = – x and Q = x
...
(2)
✞ x2
I = ✟ ( x) e
Let
Let
✠x
2
dx
2
2
✡
t , then – x dx = dt or x dx = – dt
...
(3)
Now (3) represents the equation of family of curves
...
Substituting x = 0 and
y = 1 in equation (3) we get
1 = – 1 + C
...
EXERCISE 9
...
dy
✒ 2 y ✓ sin x
dx
4
...
dy
2
✒ 2 y ✓ x log x
dx
8
...
3
...
cos x
x
7
...
MATHEMATICS
x
dy
dx
y ✁ x xy cot x ✂ 0 ( x ✄ 0) 10
...
dx
For each of the differential equations given in Exercises 13 to 15, find a particular
solution satisfying the given condition:
11
...
dy
dx
2 y tan x ✂ sin x; y ✂ 0 when x ✂
2
14
...
16
...
18
...
2
12
...
Find the equation of a curve passing through the point (0, 2) given that the sum of
the coordinates of any point on the curve exceeds the magnitude of the slope of
the tangent to the curve at that point by 5
...
(1)
dy
= e ax – bc1 sin bx ✂ b c2 cos bx ✁ ✂ c1 cos bx ✂ c2 sin bx✁ e ax ✄ a
dx
dy
= e ax [(b c2 ☎ a c1 ) cos bx ☎ ( a c2 ✆ b c1 ) sin bx]
dx
Differentiating both sides of equation (2) with respect to x, we get
or
d2y
= e ax [(b c2 ☎ a c1 ) ( ✆ b sin bx ) ☎ (a c2
dx 2
+ [(b c2 ☎ a c1 ) cos bx ☎ ( a c2
✆
✆
...
a
= e ax [( a 2 c2 ✆ 2ab c1 ✆ b 2 c2 ) sin bx ☎ (a 2 c1 ☎ 2ab c2 ✆ b 2c1 ) cos bx]
d 2 y dy
,
and y in the given differential equation, we get
dx 2 dx
Substituting the values of
L
...
S
...
H
...
Hence, the given function is a solution of the given differential equation
...
Solution Let C denote the family of circles in the second quadrant and touching the
coordinate axes
...
6)
...
(1)
(x + a)2 + (y – a)2 = a2
2
2
...
6
Substituting the value of a in equation (1), we get
2
2
2
x ✍ y y✌ ✏
✎ x ✍ y y✌ ✏ ✎
✎ x ✍ y y✌ ✏
x
y
✍
✍
✑
=
✒ y✌ ✑ 1 ✓
✒
y✌ ✑ 1 ✓✕ ✒✔
y✌ ✑ 1 ✓✕
✔
✕
✔
[xy✖ – x + x + y y✖ ]2 + [y y✖ – y – x – y y✖]2 = [x + y y✖]2
(x + y)2 y✖2 + [x + y]2 = [x + y y✖]2
(x + y)2 [(y✖)2 + 1] = [x + y y✖ ]2
or
or
or
which is the differential equation representing the given family of circles
...
Solution The given differential equation can be written as
dy
= e(3x + 4y)
dx
dy
= e3x
...
(1)
DIFFERENTIAL EQUATIONS
4y
e3 x
✁4
3
3x
– 4y
or
4 e + 3 e + 12 C = 0
Substituting x = 0 and y = 0 in (2), we get
e
or
417
=
✂C
4 + 3 + 12 C = 0 or C =
...
Example 27 Solve the differential equation
(x dy – y dx) y sin
☎
✝
✟
y✆
✞ = (y dx + x dy) x cos
x✠
☎
✝
✟
y✆
✞
...
(1)
dy
dx
✮
☎
g✝
✟
y✆
✞
...
(2)
418
MATHEMATICS
or
v x
or
x
or
Therefore
or
or
or
or
Replacing v by
v cos v ✁ v 2 sin v
dv
=
v sin v ✂ cos v
dx
(using (1) and (2))
dv
2v cos v
=
dx
v sin v ✄ cos v
✆ v sin v ☎ cos v ✝ dv 2 dx
✞
✟ = x
✠ v cos v ✡
1
✆ v sin v ☎ cos v ✝
☛ ✞✠ v cos v ✟✡ dv = 2 ☞ x dx
1
1
☞ tan v dv ✌ ☞ v dv = 2 ☞ x dx
log sec v ✍ log | v | = 2 log | x | ✎ log | C1 |
log
sec v
= log | C1 |
v x2
secv
= ± C1
v x2
...
Example 28 Solve the differential equation
(tan–1y – x) dy = (1 + y2) dx
...
(1)
DIFFERENTIAL EQUATIONS
dx
dy
Now (1) is a linear differential equation of the form
where,
P1 =
419
P1 x = Q1,
1
tan ✁ 1y
...
F = ✝ 1✆ y 2 dy
e
✞ e tan y
Thus, the solution of the given differential equation is
Therefore,
xe
tan
✟
1
y
☛ tan✡1 y ☞ tan ✠1y
e
dy ✌ C
= ✒✍
2 ✎
✏ 1✌ y ✑
...
et dt = t et – et = et (t – 1)
t
or
I = etan✢ y (tan–1y –1)
Substituting the value of I in equation (2), we get
1
✣
✥ e tan y (tan ✤1 y ✦ 1) ✧ C
✣
x = (tan ✤1y ✦ 1) ✧ C e✤ tan y
x
...
Miscellaneous Exercise on Chapter 9
1
...
2
(i)
(iii)
d2y
dy
✪ 5 x ★✭ ✩✮ ✫ 6 y ✬ log x
2
✯ dx ✰
dx
✓ d3y ✔
d4y
✱ sin ✖ 3 ✗ ✕ 0
4
dx
✙ dx ✚
3
2
★ dy ✩ ✫ 4 ★ dy ✩ ✪ 7 y ✬ sin x
✮
✭ ✮
✯ dx ✰
✯ dx ✰
(ii) ✭
420
MATHEMATICS
2
...
d2y
dx 2
2
dy
✁ xy
dx
(i) y = a ex + b e–x + x2
: x
(ii) y = ex (a cos x + b sin x)
:
d2y
dy
✄ 2 ☎ 2y ✆ 0
2
dx
dx
(iii) y = x sin 3x
:
d2y
dx 2
x2 ✁ 2 ✂ 0
9 y ✁ 6cos3 x ✂ 0
dy
✞ xy ✟ 0
dx
3
...
4
...
5
...
(iv) x2 = 2y2 log y
2
2
: (x ✝ y )
1 ✠ y2
dy
✡
☛0
...
Find the general solution of the differential equation
dx
1 ✠ x2
dy y 2 y 1
✂ 0 is
dx x 2 x 1
given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter
...
Show that the general solution of the differential equation
✌
☞✍
8
...
9
...
x
✓
x
✔
10
...
11
...
(Hint: put x – y = t)
DIFFERENTIAL EQUATIONS
12
...
✞
x ✠ dy
13
...
2
14
...
15
...
If the population of the village was
20, 000 in 1999 and 25000 in the year 2004, what will be the population of the
village in 2009?
16
...
The general solution of a differential equation of the type
(A) y e✔
P1 dy
(B) y
...
The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
(B) x ey + y2 = C
(A) x ey + x2 = C
(D) y ey + x2 = C
(C) y ex + x2 = C
422
MATHEMATICS
Summary
An equation involving derivatives of the dependent variable with respect to
independent variable (variables) is known as a differential equation
...
Degree of a differential equation is defined if it is a polynomial equation in its
derivatives
...
A function which satisfies the given differential equation is called its solution
...
To form a differential equation from a given function we differentiate the
function successively as many times as the number of arbitrary constants in
the given function and then eliminate the arbitrary constants
...
e
...
A differential equation which can be expressed in the form
✁
✁
dy
dx
f ( x, y ) or
g ( x, y ) where, f (x, y) and g(x, y) are homogenous
dx
dy
functions of degree zero is called a homogeneous differential equation
...
A differential equation of the form
Historical Note
One of the principal languages of Science is that of differential equations
...
DIFFERENTIAL EQUATIONS
423
Leibnitz was actually interested in the problem of finding a curve whose tangents
were prescribed
...
A year later he formulated the ‘method of solving the homogeneous
differential equations of the first order’
...
How surprising is it that all these methods came from a single man
and that too within 25 years of the birth of differential equations!
In the old days, what we now call the ‘solution’ of a differential equation,
was used to be referred to as ‘integral’ of the differential equation, the word
being coined by James Bernoulli (1654 - 1705) in 1690
...
It was Jules Henri Poincare
(1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the
word ‘solution’ has found its deserved place in modern terminology
...
Application to geometric problems were also considered
...
In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the
differential equation
x2 y = 2y,
which led to three types of curves, viz
...
This shows how varied the solutions of such innocent looking
differential equation can be
...
Now-a-days, this has acquired prime importance being absolutely
necessary in almost all investigations
...
– A
...
1 Introduction
In Class XI, while studying Analytical Geometry in two
dimensions, and the introduction to three dimensional
geometry, we confined to the Cartesian methods only
...
We will now use vector algebra
to three dimensional geometry
...
In this chapter, we shall study the direction cosines
and direction ratios of a line joining two points and also
discuss about the equations of lines and planes in space
under different conditions, angle between two lines, two
Leonhard Euler
planes, a line and a plane, shortest distance between two
(1707-1783)
skew lines and distance of a point from a plane
...
Nevertheless, we shall also translate
these results in the Cartesian form which, at times, presents a more clear geometric
and analytic picture of the situation
...
2 Direction Cosines and Direction Ratios of a Line
From Chapter 10, recall that if a directed line L passing through the origin makes
angles , ✂ and ✄ with x, y and z-axes, respectively, called direction angles, then cosine
of these angles, namely, cos , cos ✂ and cos ✄ are called direction cosines of the
directed line L
...
e
...
Thus, the signs of the direction cosines are reversed
...
1
Note that a given line in space can be extended in two opposite directions and so it
has two sets of direction cosines
...
These unique
direction cosines are denoted by l, m and n
...
Now take one of the directed lines from the origin and find its direction cosines as two
parallel line have same set of direction cosines
...
If l, m, n are direction cosines and a, b, c are
direction ratios of a line, then a = l, b= m and c = n, for any nonzero ✁ R
...
Let a, b, c be direction ratios of a line and let l, m and n be the direction cosines
(d
...
Then
Therefore
But
Therefore
or
l
m
=
=
a
b
l = ak, m =
l2 + m2 + n2 =
2
2
k (a + b2 + c2) =
n
✄ k (say), k being a constant
...
(1)
THREE DIMENSIONAL GEOMETRY
465
Hence, from (1), the d
...
’s of the line are
l ✁
a
a ✂b ✂c
2
2
2
,m ✁
b
a ✂b ✂c
2
2
2
,n ✁
c
a ✂ b2 ✂ c2
2
where, depending on the desired sign of k, either a positive or a negative sign is to be
taken for l, m and n
...
So, any two sets of direction ratios of a line are also proportional
...
11
...
1 Relation between the direction cosines of a line
Consider a line RS with direction cosines l, m, n
...
From P draw a perpendicular
PA on the x-axis (Fig
...
2)
...
This gives x = lr
...
Then cos ☎ ✆
Similarly,
Thus
But
Hence
Z
S
R
P (x, y , z)
O
Y
P
a
r
O
a
x
A
Fig 11
...
2
...
3 (a))
...
3
466
MATHEMATICS
Let l, m, n be the direction cosines of the line PQ and let it makes angles , ✁ and ✂
with the x, y and z-axis, respectively
...
Draw a
perpendicular from P to QS to meet at N
...
3 (b)
...
Solution Let the d
...
's of the lines be l , m, n
...
2
Example 2 If a line has direction ratios 2, – 1, – 2, determine its direction cosines
...
THREE DIMENSIONAL GEOMETRY
467
Solution We know the direction cosines of the line passing through two points
P(x1, y1, z1) and Q(x2, y2, z2) are given by
x2 x1 y2
y1 z2 z1
,
,
PQ
PQ
PQ
where
PQ =
( x2 ✄ x1 ) 2 ☎ ( y 2 ✄ y1 ) 2 ☎ ✁z 2 ✄ z1 ✂
2
Here P is (– 2, 4, – 5) and Q is (1, 2, 3)
...
Solution The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis
...
e
...
Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively
...
Solution Direction ratios of line joining A and B are
1 – 2, – 2 – 3, 3 + 4 i
...
, – 1, – 5, 7
...
e
...
It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel
to BC
...
Therefore, A, B, C are
collinear points
...
1
1
...
2
...
3
...
Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear
...
Find the direction cosines of the sides of the triangle whose vertices are
(3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2)
...
3 Equation of a Line in Space
We have studied equation of lines in two dimensions in Class XI, we shall now study
the vector and cartesian equations of a line in space
...
✁
11
...
1 Equation of a line through a given point and parallel to a given vector b
✁
Let a be the position vector of the given point
A with respect to the origin O of the
rectangular coordinate system
...
Let r be the
position vector of an arbitrary point P on the
line (Fig 11
...
✁
✄✄✄✂
✄✄✄✂
Then AP is parallel to the vector b , i
...
,
✁
Fig 11
...
✄✄✄✂
✄✄✄✂
✁
☎b
✂
✂
✂
✄✄✄✂
AP = OP – OA
But
✂
= r ✆a
Conversely, for each value of the parameter ☎, this equation gives the position
vector of a point P on the line
...
e
...
(1)
Remark If b ✞ aiˆ ✟ bjˆ ✟ ckˆ , then a, b, c are direction ratios of the line and conversely,
✂
if a, b, c are direction ratios of a line, then b ✞ aiˆ ✟ bjˆ ✟ ckˆ will be the parallel to
✁
the line
...
Derivation of cartesian form from vector form
Let the coordinates of the given point A be (x1, y1, z1) and the direction ratios of
the line be a, b, c
...
Then
✁
✁
r ✡ xiˆ ✠ yˆj ✠ zkˆ ; a ✡ x iˆ ✠ y ˆj ✠ z kˆ
and
✂
1
1
1
b ✞ a iˆ ✟ b ˆj ✟ c kˆ
Substituting these values in (1) and equating the coefficients of iˆ, ˆj and kˆ , we get
...
Eliminating the parameter
we get
x – x1
y – y1 z – z1
=
=
a
b
c
This is the Cartesian equation of the line
...
(3)
✁Note If l, m, n are the direction cosines of the line, the equation of the line is
x – x1
y – y1 z – z1
=
=
m
n
l
Example 6 Find the vector and the Cartesian equations of the line through the point
(5, 2, – 4) and which is parallel to the vector 3 iˆ ✂ 2 ˆj ✄ 8 kˆ
...
xiˆ ✂ y ˆj ✂ z kˆ = 5 iˆ ✂ 2 ˆj ✄ 4 kˆ ✂ ✞ ( 3 iˆ ✂ 2 ˆj ✄ 8 kˆ)
Therefore,
= (5 ✡ 3☛ ) ✌i ✡ (2 ✡ 2☛ ) ✟j ✡ ( ☞ 4 ☞ 8☛ ) k✠
Eliminating , we get
x ✍5
y✄2 z✂4
✆
=
3
2
✄8
which is the equation of the line in Cartesian form
...
3
...
5)
...
Therefore,
P is on the line if and only if
✝ ✝
✝ ✝
r ✄ a ✆ ✞ (b ✄ a )
Fig 11
...
or
This is the vector equation of the line
...
(1)
Derivation of cartesian form from vector form
We have
☎
☎
☎
r ✂ xiˆ ✄ y ˆj ✄ z kˆ, a ✂ x1iˆ ✄ y1 ˆj ✄ z1 kˆ and b ✂ x2iˆ ✄ y2 ˆj ✄ z2 kˆ,
Substituting these values in (1), we get
x ☛i ✞ y ✆j ✞ z k✝ ✟ x1 ☛i ✞ y1 ✆j ✞ z1 k✝ ✞ ✠ [( x2 ✡ x1 ) ☛i ✞ ( y2 ✡ y1 ) ✆j ✞ ( z2 ✡ z1 ) k✝ ]
Equating the like coefficients of iˆ, ˆj , kˆ , we get
x = x1 + (x2 – x1); y = y1 + (y2 – y1); z = z1 +
On eliminating , we obtain
x x1
y y1
z z1
x2 x1 y2 y1 z2 z1
(z2 – z1)
which is the equation of the line in Cartesian form
...
☞
☞
Solution Let a and b be the position vectors of the point A (– 1, 0, 2) and B (3, 4, 6)
...
Then the vector equation of
the line is
☎
r ✂ ✌ iˆ ✄ 2 kˆ ✄ ✍ (4 iˆ ✄ 4 ˆj ✄ 4 kˆ)
Therefore
Example 8 The Cartesian equation of a line is
x✎3 y✏5 z ✎6
✑
✑
2
4
2
Find the vector equation for the line
...
THREE DIMENSIONAL GEOMETRY
471
Thus, the required line passes through the point (– 3, 5, – 6) and is parallel to the
✁
vector 2 iˆ 4 ˆj 2 kˆ
...
4 Angle between Two Lines
Let L1 and L2 be two lines passing through the origin
and with direction ratios a1, b1, c1 and a2, b2, c2,
respectively
...
Consider the directed lines OP and OQ as
given in Fig 11
...
Let ✆ be the acute angle between
OP and OQ
...
Therefore, the
angle ✆ between them is given by
cos ✆ =
a1a2
Fig 11
...
(1)
a12 b12 c12 a22 b22 c22
The angle between the lines in terms of sin ✆ is given by
sin ✆ =
=
=
=
1 ✄ cos 2 ✝
1✄
(a1a2
b1b2 c1c2 ) 2
✞ a12 b12 c12 ✟✞ a22 b22 c22 ✟
✞ a12 b12 c12 ✟✞ a22 b22 c22 ✟ ✄ ✠ a1a2 b1b2 c1c2 ✡
2
✞ a12 b12 c12 ✟ ✞ a22 b22 c22 ✟
(a1 b2 ✄ a2 b1 )2
a12
(b1 c2 ✄ b2 c1 )2
b12
c12
a22
(c1 a2 ✄ c2 a1 )2
b22
c22
...
472
MATHEMATICS
If instead of direction ratios for the lines L1 and L2, direction cosines, namely,
l1, m1, n1 for L1 and l2, m2, n2 for L2 are given, then (1) and (2) takes the following form:
and
cos
= | l1 l2 + m1m2 + n1n2 |
sin
=
(as l12 ✁ m12 ✁ n12 ✂1 ✂ l22 ✁ m22 ✁ n22 )
2
2
2
✄l1 m2 ✆ l2 m1 ☎ ✆ ( m1 n2 ✆ m2 n1 ) ✁ ( n1 l2 ✆ n2 l1 )
...
(4)
Two lines with direction ratios a1, b1, c1 and a2, b2, c2 are
(i) perpendicular i
...
if = 90° by (1)
a1a2 + b1b2 + c1c2 = 0
(ii) parallel i
...
if = 0 by (2)
a1
b1
=
a2
b2
c1
c2
Now, we find the angle between two lines when their equations are given
...
(1)
x ✆ x2
y ✆ y2 z ✆ z2
✂
=
a2
b2
c2
...
If ✂ is the angle between them, then
cos ✂ =
3
...
1 ✆ 4
...
☞ 15 ✌
11
...
Also, if two lines in space are parallel,
then the shortest distance between them
will be the perpendicular distance, i
...
the
length of the perpendicular drawn from a
point on one line onto the other line
...
In fact,
such pair of lines are non coplanar and
are called skew lines
...
7
x, y and z-axes respectively Fig 11
...
474
MATHEMATICS
The line GE that goes diagonally across the ceiling and the line DB passes through
one corner of the ceiling directly above A and goes diagonally down the wall
...
By the shortest distance between two lines we mean the join of a point in one line
with one point on the other line so that the length of the segment so obtained is the
smallest
...
11
...
1 Distance between two skew lines
We now determine the shortest distance between two skew lines in the following way:
Let l1 and l2 be two skew lines with equations (Fig
...
8)
✁
✁
✁
✁
✁
✁
r = a1 ✂ b1
...
(2)
r = a2
and
✆
Take any point S on l1 with position vector a1 and T on l2, with position vector a 2
...
6
...
✝✝✝✁
If PQ is the shortest distance vector between
✁
l1 and l2 , then it being perpendicular to both b1 and
✁
✝✝✝✁
✁
✁
l1
P
S
b2 , the unit vector nˆ along PQ would therefore be
Fig 11
...
(3)
✝✝✝✁
Then
PQ = d nˆ
where, d is the magnitude of the shortest distance vector
...
Then
But
PQ = ST | cos ✟ |
cos ✟ =
✝✝✝✁ ✝✝✁
PQ ✡ ST
✝✝✝✝✁ ✝✝✁
| PQ | | ST |
☎
☎
d nˆ ☛ (a2 ☞ a1 )
=
d ST
✁
=
✁
✁
✁ ✁
ST b1 ✞ b2
✝✝✁
✁
✁
(since ST ✌ a2 ✍ a1 )
✁
(b1 ✞ b2 ) ✡ (a2 ✍ a1)
[From (3)]
THREE DIMENSIONAL GEOMETRY
475
Hence, the required shortest distance is
d = PQ = ST | cos |
( b1
d=
or
b2 )
...
5
...
Let the lines be given by
☎
☎
☎
...
9
...
✡
✠✠✟
Let be the angle between the vectors ST and b
...
9
where nˆ is the unit vector perpendicular to the plane of the lines l1 and l2
✠✠✟ ✟
✟
But
ST = a2 ✍ a1
...
e
...
(1)
r = iˆ ✆ ˆj ✆ ✝ (2 iˆ ✁ ˆj ✆ kˆ )
✂
r = 2 iˆ ✆ ˆj ✁ kˆ ✆ ✞ (3 iˆ ✁ 5 ˆj ✆ 2 kˆ )
and
...
( a2 ✏ a1 )
☛ ☛
| b1 ✎ b2 |
✑
| 3✏ 0 ✒ 7 |
59
✑
10
59
Example 12 Find the distance between the lines l1 and l2 given by
✂
r = iˆ ✆ 2 ˆj ✁ 4 kˆ ✆ ✝ ( 2 iˆ ✆ 3 ˆj ✆ 6 kˆ )
and
✂
r = 3 iˆ ✆ 3 ˆj ✁ 5 kˆ ✆ ✞ ( 2 iˆ ✆ 3 ˆj ✆ 6 kˆ )
THREE DIMENSIONAL GEOMETRY
Solution The two lines are parallel (Why? ) We have
✄
477
✄
✁
a1 = iˆ 2 ˆj ✂ 4 kˆ , a2 = 3 iˆ 3 ˆj ✂ 5 kˆ and b = 2 iˆ 3 ˆj 6 kˆ
Therefore, the distance between the lines is given by
✆
✆
iˆ ˆj kˆ
2 3 6
2 1 ✂1
✆
b ☎ (a2 ✂ a1 )
✆
d=
=
|b |
4
or
=
9
36
| ✂ 9 iˆ 14 ˆj ✂ 4 kˆ |
293
293
✝
✝
7
49
49
EXERCISE 11
...
Show that the three lines with direction cosines
12 ✂3 ✂4
4 12 3
3 ✂4 12
,
,
;
,
,
;
,
,
are mutually perpendicular
...
Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the
line through the points (0, 3, 2) and (3, 5, 6)
...
Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line
through the points (– 1, – 2, 1), (1, 2, 5)
...
Find the equation of the line which passes through the point (1, 2, 3) and is
parallel to the vector 3 iˆ 2 ˆj ✂ 2 kˆ
...
Find the equation of the line in vector and in cartesian form that passes through
the point with position vector 2 iˆ ✂ j 4 kˆ and is in the direction iˆ 2 ˆj ✂ kˆ
...
Find the cartesian equation of the line which passes through the point (– 2, 4, – 5)
and parallel to the line given by
x✞3 y✟4 z✞8
...
Write its vector form
...
Find the vector and the cartesian equations of the lines that passes through the
origin and (5, – 2, 3)
...
The cartesian equation of a line is
478
MATHEMATICS
9
...
10
...
Find the angle between the following pair of lines:
x ✁ 2 y ✁1 z ✂ 3
x✂ 2 y✁ 4 z ✁5
and
(i)
2
5
8
4
✁3
✁1
(ii) r
x y z
x✝5 y✝2 z✝3
✞ ✞ and
✞
✞
2 2 1
4
1
8
1✁ x
12
...
Show that the lines
7 y ✁ 14
2p
z✁3
2
6✁ z
are at right angles
...
14
...
Find the shortest distance between the lines
x✝3 y✝5 z✝7
x ✂1 y ✂1 z ✂1
✞
✞
and
1
1
7
1
✝2
✁6
16
...
Find the shortest distance between the lines whose vector equations are
☎
r (1 ✁ t ) iˆ ✂ (t ✁ 2) ˆj ✂ (3 ✁ 2 t ) kˆ and
☎
r
( s ✂ 1) iˆ ✂ (2s ✁ 1) ˆj ✁ (2s ✂ 1) kˆ
THREE DIMENSIONAL GEOMETRY
479
11
...
e
...
(ii) it passes through a point and is perpendicular to a given direction
...
Now we shall find vector and Cartesian equations of the planes
...
6
...
Fig 11
...
If ON is the normal from the origin to the plane, and nˆ is the unit normal vector
✂✂✂✁
✄✄✄☎
❩
along ON
...
Let P be any
✂✂✂✁
point on the plane
...
✂✂✂✁ ✂✂✂✁
Therefore, NP ✆ ON = 0
P✏ ①✑✒✑✱ ✮
r
❞ ◆
❖
...
10
(d
0)
☛ ☛
☛ ☛
r n =d
(as n ✌ n ✎ 1)
i
...
,
This is the vector form of the equation of the plane
...
Let P(x, y, z) be any point on the plane
...
Then
nˆ = l iˆ ✟ m ˆj ✟ n kˆ
480
MATHEMATICS
Therefore, (2) gives
( x iˆ y ˆj z kˆ) ✁ (l iˆ m ˆj n kˆ) ✂ d
i
...
,
lx + my + nz = d
This is the cartesian equation of the plane in the normal form
...
(3)
☎
ˆ
✄Note Equation (3) shows that if r ✁ (a i
b ˆj c kˆ) = d is the vector equation
of a plane, then ax + by + cz = d is the Cartesian equation of the plane, where a, b
and c are the direction ratios of the normal to the plane
...
✠
Solution Let n ✟ 2 iˆ ✞ 3 ˆj ✝ 4 kˆ
...
Also find its
☎
2 iˆ ✆ 3 ˆj 4 kˆ
2 iˆ ✆ 3 ˆj 4 kˆ
n
nˆ ✂ ☎ =
✂
| n|
4 9 16
29
Hence, the required equation of the plane is
☎ ✡ 2 ˆ
i
✍ 29
r ✁☞
✆3 ˆ
j
4 ˆ☛
6
k✌✂
29 ✎
29
29
Example 14 Find the direction cosines of the unit vector perpendicular to the plane
☎
r ✁ (6 iˆ ✆ 3 ˆj ✆ 2 kˆ) 1 = 0 passing through the origin
...
(1)
| ✆ 6 iˆ 3 ˆj 2 kˆ | = 36 ✝ 9 ✝ 4 ✟ 7
Now
Therefore, dividing both sides of (1) by 7, we get
☎ ✡ 6ˆ
i
✍ 7
r ✁☞✆
3 ˆ
j
7
1
2 ˆ☛
k✌ =
7
7 ✎
✠
which is the equation of the plane in the form r ✏ nˆ ✟ d
...
Hence, the direction cosines of nˆ are ✆ 6 , 3 , 2
...
Solution Since the direction ratios of the normal to the plane are 2, –3, 4; the direction
cosines of it are
2
2 ✁ ( 3) ✁ 4
2
2
2
,
3
2 ✁ ( 3) ✁ 4
2
2
,
2
4
2 ✁ ( 3) ✁ 4
2
2
2
, i
...
,
2
✂3 4
,
,
29 29 29
Hence, dividing the equation 2x – 3y + 4z – 6 = 0 i
...
, 2x – 3y + 4z = 6 throughout by
29 , we get
✂3
2
4
6
x ✄
y ✄
z ☎
29
29
29
29
This is of the form lx + my + nz = d, where d is the distance of the plane from the
origin
...
Example 16 Find the coordinates of the foot of the perpendicular drawn from the
origin to the plane 2x – 3y + 4z – 6 = 0
...
11)
...
P(x1 , y1 , z1)
Writing the equation of the plane in the normal
form, we have
2
29
x✂
3
29
y✄
4
29
6
z☎
O
29
✂3
2
4
,
,
are the direction X
29 29 29
cosines of the OP
...
11
Since d
...
’s and direction ratios of a line are proportional, we have
x1
=
2
29
i
...
,
x1 =
y1
✆
3
29
2k
29
, y1 =
z1
=k
4
29
✂3k ,
4k
z1 ☎
29
29
482
MATHEMATICS
Substituting these in the equation of the plane, we get k =
6
29
...
☎
✆ 29 29 29 ✝
Hence, the foot of the perpendicular is ✄
✞Note If d is the distance from the origin and l, m, n are the direction cosines of
the normal to the plane through the origin, then the foot of the perpendicular is
(ld, md, nd)
...
6
...
12)
...
Fig 11
...
(Fig 11
...
Then the point P lies in the plane if and only if
☛☛☛✡
✠✟
☛☛☛✡ ✠✟
AP is perpendicular to N
...
e
...
N = 0
...
Therefore, ( r a ) N 0
… (1)
This is the vector equation of the plane
...
Then,
Fig 11
...
e
...
THREE DIMENSIONAL GEOMETRY
483
☎
Solution We have the position vector of point (5, 2, – 4) as a ✁ 5 iˆ ✂ 2 ˆj ✄ 4kˆ and the
✆
☎
normal vector N perpendicular to the plane as N = 2 iˆ +3 ˆj ✄ kˆ
☎
☎
☎
Therefore, the vector equation of the plane is given by (r ✄ a )
...
(1)
Transforming (1) into Cartesian form, we have
[( x – 5) iˆ ✂ ( y ✄ 2) ˆj ✂ ( z ✂ 4) kˆ] ✝ (2 iˆ ✂ 3 ˆj ✄ kˆ ) ✁ 0
2( x ✄ 5) ✂ 3( y ✄ 2) ✄ 1( z ✂ 4) ✁ 0
or
i
...
2x + 3y – z = 20
which is the cartesian equation of the plane
...
6
...
14)
...
14
✞✞✞
✆
✟✟✟☎
✟✟✟☎
✟✟✟☎
The vectors RS and RT are in the given plane
...
Let r be the position vector
of any point P in the plane
...
Note Why was it necessary to say that the three points
had to be non collinear? If the three points were on the same
line, then there will be many planes that will contain them
(Fig 11
...
✁
These planes will resemble the pages of a book where the
line containing the points R, S and T are members in the binding
of the book
...
15
Cartesian form
Let (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be the coordinates of the points R, S and T
respectively
...
Then
✂✂✂
RP = (x – x1) iˆ + (y – y1) ˆj + (z – z1) kˆ
✄✄✄☎
RS = (x2 – x1) iˆ + (y2 – y1) ˆj + (z2 – z1) kˆ
✂✂✂
RT = (x3 – x1) iˆ + (y3 – y1) ˆj + (z3 – z1) kˆ
Substituting these values in equation (1) of the vector form and expressing it in the
form of a determinant, we have
x x1
y y1
z z1
x2
x1
y2
y1
z2 z1
x3
x1
y3
y1
z3 z1
0
which is the equation of the plane in Cartesian form passing through three non collinear
points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3)
...
Solution Let a ✆ 2 iˆ ✝ 5 ˆj ✞ 3 kˆ , b ✆ ✞ 2 iˆ ✞ 3 ˆj ✝ 5 kˆ , c ✆ 5 iˆ ✝ 3 ˆj ✞ 3 kˆ
☎
☎
Then the vector equation of the plane passing through a☎ , b and c and is
given by
✂✂✂ ✂✂✂
(r ✞ a ) ✟ (RS✠ RT) = 0 (Why?)
or
(r ✞ a ) ✟ [(b ✞ a ) ✠ (c ✞ a ) ] = 0
i
...
[r ✞ (2 iˆ ✝ 5 ˆj ✞ 3 kˆ)] ✟ [(✞4 iˆ ✞ 8 ˆj ✝ 8 kˆ) ✠ (3 iˆ ✞ 2 ˆj )] ✆ 0
THREE DIMENSIONAL GEOMETRY
485
11
...
4 Intercept form of the equation of a plane
In this section, we shall deduce the equation of a plane in terms of the intercepts made
by the plane on the coordinate axes
...
(1)
Ax + By + Cz + D = 0 (D 0)
Let the plane make intercepts a, b, c on x, y and z axes, respectively (Fig 11
...
Hence, the plane meets x, y and z-axes at (a, 0, 0),
(0, b, 0), (0, 0, c), respectively
...
16
x y z
✄ ✄ =1
a b c
which is the required equation of the plane in the intercept form
...
(1)
Example 19 Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and
z-axis respectively
...
(1)
a b c
Here
a = 2, b = 3, c = 4
...
2 3 4
11
...
5 Plane passing through the intersection
of two given planes
Let ✝ 1 and ✝ 2 be two planes with equations
✟
✟
r ✞ nˆ1 = d1 and r ✞ nˆ2 = d2 respectively
...
17)
...
17
486
MATHEMATICS
✁
If t is the position vector of a point on the line, then
✁
✁
t nˆ1 = d1 and t nˆ2 = d2
Therefore, for all real values of ✂, we have
✁
t ( nˆ1 ✄ ☎ nˆ2 ) = d1 ✄ ☎ d 2
✆
Since t is arbitrary, it satisfies for any point on the line
...
e
...
(1)
Cartesian form
In Cartesian system, let
✠
n1 = A1 iˆ ✡ B2 ˆj ✡ C1 kˆ
✠
n2 = A 2 iˆ ✡ B2 ˆj ✡ C2 kˆ
✟
r = xiˆ ✡ y ˆj ✡ z kˆ
and
Then (1) becomes
x (A1 + ✂A2) + y (B1 + ✂B2) + z (C1 + ✂C2) = d1 + ✂d2
or
(A1 x + B1 y + C1z – d1) + ✂ (A2 x + B2 y + C2 z – d2) = 0
...
Example 20 Find the vector equation of the plane passing through the intersection of
✟
✟
the planes r ☛ (iˆ ✡ ˆj ✡ kˆ) ☞ 6 and r ☛ (2iˆ ✡ 3 ˆj ✡ 4kˆ) ☞ ✌ 5, and the point (1, 1, 1)
...
… (1)
THREE DIMENSIONAL GEOMETRY
Taking
487
✂
r xiˆ ✁ y ˆj ✁ z kˆ , we get
( xiˆ ✁ y ˆj ✁ z kˆ) ✄[(1✁ 2☎ ) iˆ ✁ (1✁ 3☎ ) ˆj ✁ (1✁ 4☎ )kˆ] 6 ✆ 5☎
or
(1 + 2✝ ) x + (1 + 3✝) y + (1 + 4✝) z = 6 – 5✝
or
(x + y + z – 6 ) + ✝ (2x + 3y + 4 z + 5) = 0
...
e
...
11
...
(1)
✂
r = a2 ✁ ✜b2
and
✢
...
The line (2) passes through the point, say B with position vector a2 and is parallel
✂
to b2
...
i
...
✣✣✣✂ ✂
✂
✂
✂
✂ ✂
AB
...
488
MATHEMATICS
✁
✁
Let a1, b1, c1 and a2, b2, c2 be the direction ratios of b1 and b2 , respectively
...
In the cartesian form,
it can be expressed as
x2 ☞ x1
y2 ☞ y1
z2 ☞ z1
a1
b2
c1
a2
b2
c2
✌0
...
–1
2
5
–3
1
5
Solution Here, x1 = – 3, y1 = 1, z1 = 5, a1 = – 3, b1 = 1, c1 = 5
x2 = – 1, y2 = 2, z2 = 5, a2 = –1, b2 = 2, c2 = 5
Now, consider the determinant
x2 ☞ x1
y2 ☞ y1
z2 ☞ z1
a1
b1
c1
a2
b2
c2
2 1 0
✌ ☞3 1 5 ✌ 0
☞1 2 5
Therefore, lines are coplanar
...
8 Angle between Two Planes
Definition 2 The angle between two planes is defined as the angle between their
normals (Fig 11
...
Observe that if ✍ is an angle between the two planes, then so
is 180 – ✍ (Fig 11
...
We shall take the acute angle as the angles between
two planes
...
18
THREE DIMENSIONAL GEOMETRY
489
✁
✁
If n1 and n2 are normals to the planes and be the angle between the planes
✁ ✁
✁ ✁
r ✂ n1 = d1 and r
...
Then is the angle between the normals to the planes drawn from some common
point
...
n2 = 0 and parallel if
✆
✆
n1 is parallel to n2
...
Therefore,
cos
=
A1 A 2
A12
B12
B1 B 2
C12
C1 C2
A 22 B 22 C22
☎ Note
1
...
2
...
A1
B
C
✝ 1 ✝ 1
...
Solution The angle between two planes is the angle between their normals
...
Solution Comparing the given equations of the planes with the equations
A1 x + B1 y + C1 z + D1 = 0 and A2 x + B2 y + C2 z + D2 = 0
We get
A1 = 3, B1 = – 6, C1 = 2
A2 = 2, B2 = 2, C2 = – 2
cos
=
Therefore,
3 ✄ 2 ☎ (✆6) (2) ☎ (2) (✆2)
=
✁32 ☎ (✆ 6) 2 ☎ (✆2)2 ✂
✁ 22 ☎ 22 ☎ (✆2) 2 ✂
5
5 3
✆ 10
✝
✝
21
7✄2 3
7 3
✞5 3✟
= cos-1 ✠✠
✡✡
☛ 21 ☞
11
...
19)
...
19
Consider a plane ✍2 through P parallel to the plane ✍1
...
Hence, its equation is ( r ✑ a ) ✒ nˆ ✓ 0
✏
✏
r ✎ nˆ = a ✎ nˆ
i
...
,
✌
Thus, the distance ON✔ of this plane from the origin is | a ✒ nˆ |
...
11
...
e
...
We may establish the similar results for (Fig 11
...
✁ Note
1
...
to the plane, then the perpendicular distance is
✡☛
|N|
2
...
Cartesian form
✞
Let P(x1, y1, z1) be the given point with position vector a and
Ax + By + Cz = D
be the Cartesian equation of the given plane
...
Therefore, the distance of the point (2, 5, – 3) from the given plane is
| (2 iˆ ☞ 5 ˆj ✌ 3 kˆ) ✂ (6 iˆ ✌ 3 ˆj ☞ 2 kˆ) ✌ 4|
| 12 ✌ 15 ✌ 6 ✌ 4 | 13
✄
=
ˆ
ˆ
ˆ
| 6 i ✌3 j ☞ 2 k |
7
36 ☞ 9 ☞ 4
492
MATHEMATICS
11
...
20)
...
Then the angle ✆ between the line and the
r ✄a✂
Fig 11
...
e
...
e
...
Solution Let ✆ be the angle between the line and the normal to the plane
...
3
1
...
(a) z = 2
(b) x + y + z = 1
(c) 2x + 3y – z = 5
(d) 5y + 8 = 0
2
...
3
...
In the following cases, find the coordinates of the foot of the perpendicular
drawn from the origin
...
Find the vector and cartesian equations of the planes
(a) that passes through the point (1, 0, – 2) and the normal to the plane is
iˆ ✄ ˆj ☎ kˆ
...
6
...
(a) (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)
(b) (1, 1, 0), (1, 2, 1), (– 2, 2, – 1)
7
...
8
...
9
...
10
...
( 2 iˆ ✄ 2 ˆj ☎ 3 kˆ ) ✆ 7 , r
...
11
...
494
MATHEMATICS
12
...
13
...
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0
and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0
and 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0
and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0
and y + z – 4 = 0
14
...
Point
Plane
(a) (0, 0, 0)
3x – 4y + 12 z = 3
(b) (3, – 2, 1)
2x – y + 2z + 3 = 0
(c) (2, 3, – 5)
x + 2y – 2z = 9
(d) (– 6, 0, 0)
2x – 3y + 6z – 2 = 0
Miscellaneous Examples
Example 26 A line makes angles
cos2
✆
, ✝, ✞ and
✆
✟
with the diagonals of a cube, prove that
+ cos2 ✝ + cos2 ✞ + cos2 ✟ =
4
3
Solution A cube is a rectangular parallelopiped having equal length, breadth and height
...
(Fig 11
...
The direction cosines of the diagonal OE which
❈✡☛☞ ☛☞ ✌✵
is the line joining two points O and E are
❋✡☛☞ ✌ ☞ ✌✵
a✂0
a2
✁
a2
✁
a2
,
a✂0
a2
✁
a2 ✁ a2
,
a✂0
a2
✁
✡✌☞ ☛☞ ✌✵ ●
❊✍✑✎✑✎✑✏
a2 ✁ a2
❖
i
...
,
1
3
,
1
3
,
1
3
❨
❇✡☛☞ ✌ ☞ ☛✵
❆✭❛✱ ✠✱ ✠✮ ❉✡✌☞ ✌ ☞ ☛✵
❳
Fig 11
...
3
3
3
3
3
Let l, m, n be the direction cosines of the given line which makes angles
with OE, AF, BG, CD, respectively
...
Solution The equation of the plane containing the given point is
A (x – 1) + B(y + 1) + C (z – 2) = 0
...
Hence, the required
equation is
– 5C (x – 1) + 4 C (y + 1) + C(z – 2) = 0
i
...
5x – 4y – z = 7
Example 28 Find the distance between the point P(6, 5, 9) and the plane determined
by the points A (3, – 1, 2), B (5, 2, 4) and C(– 1, – 1, 6)
...
D is the foot of the perpendicular
drawn from a point P to the plane
...
496
MATHEMATICS
✂✂✂✁
✂✂✂✁
Hence, PD = the dot product of AP with the unit vector along AB
✂✂✂✁
✂✂✂✁
AC
...
3 i ✆ 4 j ✝ 3 k
34
3 34
17
Alternatively, find the equation of the plane passing through A, B and C and then
compute the distance of the point P from the plane
...
b–c
b
b+c
✍–✎
✍
✍+✎
b ✆c✆ a ✝ d
z2 ✆ z1
c1
✟✆✠
=
c2
✡✆☛
b ✆ a b ✝c ✆ a ✆ d
✟
✡
✟✝✠
✡✝☛
THREE DIMENSIONAL GEOMETRY
497
Adding third column to the first column, we get
b
2
a
✂
☎
a b ✁c
b
✂
☎
a
d
✂ ✁✄
☎✁ ✆
=0
Since the first and second columns are identical
...
Example 30 Find the coordinates of the point where the line through the points
A (3, 4, 1) and B (5, 1, 6) crosses the XY-plane
...
(1)
r = 3 iˆ ✁ 4 ˆj ✁ kˆ ✁ ✞ ( 2 iˆ 3 ˆj ✁ 5 kˆ )
Let P be the point where the line AB crosses the XY-plane
...
i
...
This point must satisfy the equation (1)
...
e
...
✍5 5
☛
Hence, the coordinates of the required point are ✌
Miscellaneous Exercise on Chapter 11
1
...
2
...
Find the angle between the lines whose direction ratios are a, b, c and
b – c, c – a, a – b
...
Find the equation of a line parallel to x-axis and passing through the origin
...
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and
(2, 9, 2) respectively, then find the angle between the lines AB and CD
...
If the lines
x
1
✁
y
2
✁
z 3
x 1 y 1 z 6
are perpendicular,
and
✁
✁
2
3k
1
5
3
2k
find the value of k
...
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to
✆
the plane r
...
8
...
✠
9
...
11
...
13
...
✠
and r ✟ ☛ 4 iˆ ☛ kˆ ✞ ☞ (3 iˆ ☛ 2 ˆj ☛ 2 kˆ)
...
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1)
crosses the ZX-plane
...
Find the equation of the plane passing through the point (– 1, 3, 2) and perpendicular
to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0
...
15
...
16
...
17
...
THREE DIMENSIONAL GEOMETRY
499
18
...
19
...
20
...
✞
✞
✞
and
=
3
7
3
8
✝ 16
✝5
21
...
✂ 2 ✂ 2
2
a
b
c
p2
Choose the correct answer in Exercises 22 and 23
...
Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is
the origin, then
(A) 2 units
(B) 4 units
(C) 8 units
(D)
2
units
29
23
...
5y + 10z = 6 are
(A) Perpendicular
(B) Parallel
(C) intersect y-axis
5✡
✠
(D) passes through ☛ 0,0, ☞
4✍
✌
Summary
✎ Direction cosines of a line are the cosines of the angles made by the line
with the positive directions of the coordinate axes
...
✎ Direction cosines of a line joining two points P(x1, y1, z1) and Q(x2, y2, z2) are
x2 ✏ x1 y2 ✏ y1 z2 ✏ z1
,
,
PQ
PQ
PQ
where PQ =
( x2 ✏ x1 ) 2 ✓ ( y2 ✏ y1 ) 2 ✓ ✑z 2 ✏ z1 ✒
2
✎ Direction ratios of a line are the numbers which are proportional to the
direction cosines of a line
...
They lie in different planes
...
If l1, m1, n1 and l2, m2, n2 are the direction cosines of two lines; and ✁ is the
acute angle between the two lines; then
cos✁ = | l1l2 + m1m2 + n1n2 |
If a1, b1, c1 and a2, b2, c2 are the direction ratios of two lines and ✁ is the
acute angle between the two lines; then
cos ✁ =
a1 a2
a12
b12
b1 b2
c12
a22
c1 c2
b22
c22
the given point whose position
✂ Vector equation of a line that passes through
✄ ✝ ✝
✝
✄
vector is a and parallel to a given vector b is r ☎ a ✆ b
...
✂ Cartesian equation of a line that passes through two points (x1, y1, z1) and
(x2, y2, z2) is
x ✠ x1
y ✠ y1
z ✠ z1
...
✂ If
THREE DIMENSIONAL GEOMETRY
501
✂ Shortest distance between two skew lines is the line segment perpendicular
to both the lines
...
✂ Equation of a plane which is at a distance of d from the origin and the direction
cosines of the normal to the plane as l, m, n is lx + my + nz = d
...
N 0
...
[ (b a ) ☎ ( c a ) ] ✆ 0
✂ Equation of a plane that cuts the coordinates axes at (a, 0, 0), (0, b, 0) and
(0, 0, c) is
x
y z
✞ ✞ ✝1
a
b c
✂ Vector equation of a plane that passes through the intersection of
✡ ✡
✡ ✡
✡ ✡
✡
planes r ✟ n1 ✠ d1 and r ✟ n2 ✠ d 2 is r ✟ (n1 ☛ ☞ n2 ) ✠ d1 ☛ ☞ d 2 , where ✌ is any
nonzero constant
...
✁
✁
✁ ✁
✁ ✁
✂ Two planes r ✆ a1 ✍ ✎ b1 and r ✆ a2 ✍ ✏ b2 are coplanar if
✁ ✁ ✁ ✁
(a2 a1 ) ✑ (b1 ☎ b2 ) = 0
✂ Two planes a1 x + b1 y + c1 z + d1 = 0 and a2 x + b2 y + c2 z + d2 = 0 are
x2 ✒ x1 y2 ✒ y1 z 2 ✒ z1
coplanar if
a1
a2
b1
b2
c1
c2
= 0
...
| n1 | | n2 |
✁
✁ ✁
✄
The
angle
between
the
line
and the plane r ✔ nˆ ✕ d is
✖
✍
✎
r
a
b
✆
✂
✁
b ✑ nˆ
sin ✗ ✆ ✁
| b | | nˆ |
THREE DIMENSIONAL GEOMETRY
503
✂ The angle between the planes A1x + B1y + C1z + D1 = 0 and
A2 x + B2 y + C2 z + D2 = 0 is given by
cos
=
A1 A 2 ✁ B1 B2 ✁ C1 C2
A12 ✁ B12 ✁ C12
A 22 ✁ B22 ✁ C 22
✄
✄
of a point whose position vector is a from the plane r ☎ nˆ ✆ d is
✂ The distance
✄
| d ✝ a ☎ nˆ |
✂ The distance from a point (x1, y1, z1) to the plane Ax + By + Cz + D = 0 is
Ax1 ✞ By1 ✞ Cz1 ✞ D
A 2 ✞ B2 ✞ C 2
...
– G
...
1 Introduction
In earlier classes, we have discussed systems of linear
equations and their applications in day to day problems
...
Many applications in mathematics
involve systems of inequalities/equations
...
He has Rs 50,000 to invest and has storage space
of at most 60 pieces
...
He estimates that from the sale of one table, he
L
...
He wants to know how many tables and chairs he should buy
from the available money so as to maximise his total profit, assuming that he can sell all
the items which he buys
...
Thus, an optimisation
problem may involve finding maximum profit, minimum cost, or minimum use of
resources etc
...
The above stated optimisation problem is an example of linear programming
problem
...
In this chapter, we shall study some linear programming problems and their solutions
by graphical method only, though there are many other methods also to solve such
problems
...
2 Linear Programming Problem and its Mathematical Formulation
We begin our discussion with the above example of furniture dealer which will further
lead to a mathematical formulation of the problem in two variables
...
Further he would earn different profits by following different investment
strategies
...
, his investment is
limited to a maximum of Rs 50,000 and so is his storage space which is for a
maximum of 60 pieces
...
e
...
His profit in this case will be Rs (250 × 20), i
...
, Rs 5000
...
With his capital of Rs 50,000,
he can buy 50000 ÷ 500, i
...
100 chairs
...
Therefore, he
is forced to buy only 60 chairs which will give him a total profit of Rs (60 × 75), i
...
,
Rs 4500
...
Total profit in this case would be
Rs (10 × 250 + 50 × 75), i
...
, Rs 6250 and so on
...
Now the problem is : How should he invest his money in order to get maximum
profit? To answer this question, let us try to formulate the problem mathematically
...
2
...
Obviously, x and y must be non-negative, i
...
,
x 0✁
✂ (Non-negative constraints)
y 0✄
...
(2)
The dealer is constrained by the maximum amount he can invest (Here it is
Rs 50,000) and by the maximum number of items he can store (Here it is 60)
...
(3)
and
x + y ☎ 60 (storage constraint)
...
(5)
Mathematically, the given problems now reduces to:
Maximise Z = 250x + 75y
subject to the constraints:
5x + y ✂ 100
x + y ✂ 60
x ✄ 0, y ✄ 0
So, we have to maximise the linear function Z subject to certain conditions determined
by a set of linear inequalities with variables as non-negative
...
Such problems
are called Linear Programming Problems
...
The term linear implies that all the mathematical relations used in the problem are
linear relations while the term programming refers to the method of determining a
particular programme or plan of action
...
In the above example, Z = 250x + 75y is a linear objective function
...
Constraints The linear inequalities or equations or restrictions on the variables of a
linear programming problem are called constraints
...
In the above example, the set of inequalities (1) to (4)
are constraints
...
Linear programming
problems are special type of optimisation problems
...
We will now discuss how to find solutions to a linear programming problem
...
12
...
2 Graphical method of solving linear programming problems
In Class XI, we have learnt how to graph a system of linear inequalities involving two
variables x and y and to find its solutions graphically
...
2
...
Let us graph the constraints stated as linear inequalities:
5x + y ✂ 100
...
(2)
x✄0
...
(4)
The graph of this system (shaded region) consists of the points common to all half
planes determined by the inequalities (1) to (4) (Fig 12
...
Each point in this region
represents a feasible choice open to the dealer for investing in tables and chairs
...
Every point of this
region is called a feasible solution to the problem
...
In Fig 12
...
The region other than feasible region is called an
infeasible region
...
In
Fig 12
...
For example, the point (10, 50) is a feasible
solution of the problem and so are the points
(0, 60), (20, 0) etc
...
For example,
the point (25, 40) is an infeasible solution of
the problem
...
1
508
MATHEMATICS
Optimal (feasible) solution: Any point in the feasible region that gives the optimal
value (maximum or minimum) of the objective function is called an optimal solution
...
To handle this situation, we use the following theorems which are
fundamental in solving linear programming problems
...
Theorem 1 Let R be the feasible region (convex polygon) for a linear programming
problem and let Z = ax + by be the objective function
...
Theorem 2 Let R be the feasible region for a linear programming problem, and let
Z = ax + by be the objective function
...
Remark If R is unbounded, then a maximum or a minimum value of the objective
function may not exist
...
(By Theorem 1)
...
Let us now compute the values of Z at these points
...
** A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a
circle
...
Unbounded means that the feasible region does extend
indefinitely in any direction
...
e
...
This method of solving linear programming problem is referred as Corner Point
Method
...
Find the feasible region of the linear programming problem and determine its
corner points (vertices) either by inspection or by solving the two equations of
the lines intersecting at that point
...
Evaluate the objective function Z = ax + by at each corner point
...
3
...
(ii) In case, the feasible region is unbounded, we have:
4
...
Otherwise, Z
has no maximum value
...
Otherwise, Z
has no minimum value
...
(1)
subject to the constraints:
x + y ✂ 50
...
(3)
x ✄ 0, y ✄ 0
...
2 is the feasible region determined by the system
of constraints (2) to (4)
...
So,
we now use Corner Point Method to determine the maximum value of Z
...
Now we evaluate Z at each corner point
...
2
Hence, maximum value of Z is 120 at the point (30, 0)
...
(1)
...
(3)
...
3 is the feasible region ABC determined by the
system of constraints (2) to (4), which is bounded
...
3
2500
2300 ☎
3000
Minimum
LINEAR PROGRAMMING
511
A, B and C are (0,5), (4,3) and (0,6) respectively
...
Hence, minimum value of Z is 2300 attained at the point (4, 3)
Example 3 Solve the following problem graphically:
Minimise and Maximise Z = 3x + 9y
subject to the constraints:
x + 3y ✂ 60
x + y ✄ 10
x✂y
...
...
x ✄ 0, y ✄ 0
(1)
(2)
(3)
(4)
...
The feasible region ABCD is shown in the Fig 12
...
Note that the region is
bounded
...
Corner
Point
A (0, 10)
B (5, 5)
C (15, 15)
D (0, 20)
Corresponding value of
Z = 3x + 9y
90
60 ☎
180
☎
180
}
Minimum
Maximum
(Multiple
optimal
solutions)
Fig 12
...
From the table, we find that
the minimum value of Z is 60 at the point B (5, 5) of the feasible region
...
Remark Observe that in the above example, the problem has multiple optimal solutions
at the corner points C and D, i
...
the both points produce same maximum value 180
...
Same is also true in the case if the
two points produce same minimum value
...
subject to the constraints:
...
...
(1)
(2)
(3)
(4)
(5)
Solution First of all, let us graph the feasible region of the system of inequalities (2) to
(5)
...
5
...
We now evaluate Z at the corner points
...
5
From this table, we find that – 300 is the smallest value of Z at the corner point
(6, 0)
...
But here we see that the feasible region is unbounded
...
To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method
...
e
...
If it has common points, then –300 will not be the minimum value of Z
...
LINEAR PROGRAMMING
513
As shown in the Fig 12
...
Therefore, Z = –50 x + 20 y
has no minimum value subject to the given constraints
...
(Why?)
Example 5 Minimise Z = 3x + 2y
subject to the constraints:
x+y✄8
...
(2)
x ✄ 0, y ✄ 0
...
6)
...
6, you can see that
there is no point satisfying all the
constraints simultaneously
...
Remarks From the examples which we
have discussed so far, we notice some
general features of linear programming
problems:
(i) The feasible region is always a
convex region
...
6
(ii) The maximum (or minimum)
solution of the objective function occurs at the vertex (corner) of the feasible
region
...
EXERCISE 12
...
Maximise Z = 3x + 4y
subject to the constraints : x + y ✂ 4, x ✄ 0, y ✄ 0
...
Minimise Z = – 3x + 4 y
subject to x + 2y ✂ 8, 3x + 2y ✂ 12, x ✄ 0, y ✄ 0
...
Maximise Z = 5x + 3y
subject to 3x + 5y ✂ 15, 5x + 2y ✂ 10, x ✄ 0, y ✄ 0
...
Minimise Z = 3x + 5y
such that x + 3y ✄ 3, x + y ✄ 2, x, y ✄ 0
...
Maximise Z = 3x + 2y
subject to x + 2y ✂ 10, 3x + y ✂ 15, x, y ✄ 0
...
Minimise Z = x + 2y
subject to 2x + y ✄ 3, x + 2y ✄ 6, x, y ✄ 0
...
7
...
8
...
9
...
10
...
12
...
Manufacturing problems In these problems, we determine the number of units
of different products which should be produced and sold by a firm
when each product requires a fixed manpower, machine hours, labour hour per
unit of product, warehouse space per unit of the output etc
...
2
...
3
...
LINEAR PROGRAMMING
515
Let us now solve some of these types of linear programming problems:
Example 6 (Diet problem): A dietician wishes to mix two types of foods in such a
way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10
units of vitamin C
...
Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C
...
Formulate
this problem as a linear programming problem to minimise the cost of such a mixture
...
Clearly, x ✄ 0,
y ✄ 0
...
(1)
2x + y ✄ 8
...
(3)
subject to the constraints:
x, y ✄ 0
...
The feasible region determined by the
system is shown in the Fig 12
...
Here again, observe that the feasible region is
unbounded
...
Corner Point
(0,8)
(2,4)
(10,0)
Z = 50x + 70y
560
380 ☎
500
Minimum
Fig 12
...
Can we say
that the minimum value of Z is 380? Remember that the feasible region is unbounded
...
e
...
From the Fig 12
...
Thus, the minimum value of Z is 380 attained at the point (2, 4)
...
Example 7 (Allocation problem) A cooperative society of farmers has 50 hectare
of land to grow two crops X and Y
...
To control weeds, a liquid herbicide
has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare
...
How much land should be allocated
to each crop so as to maximise the total profit of the society?
Solution Let x hectare of land be allocated to crop X and y hectare to crop Y
...
Profit per hectare on crop X = Rs 10500
Profit per hectare on crop Y = Rs 9000
Therefore, total profit
= Rs (10500x + 9000y)
LINEAR PROGRAMMING
517
The mathematical formulation of the problem is as follows:
Maximise
Z = 10500 x + 9000 y
subject to the constraints:
x + y ✂ 50 (constraint related to land)
...
e
...
(2)
(non negative constraint)
...
The feasible region
OABC is shown (shaded) in the Fig 12
...
Observe that the feasible region is bounded
...
Let us evaluate the objective function Z = 10500 x + 9000y at
these vertices to find which one gives the maximum profit
...
8
Hence, the society will get the maximum profit of Rs 4,95,000 by allocating 30
hectares for crop X and 20 hectares for crop Y
...
Each piece of Model A requires 9 labour hours for fabricating
and 1 labour hour for finishing
...
For fabricating and finishing, the maximum
labour hours available are 180 and 30 respectively
...
How many
pieces of Model A and Model B should be manufactured per week to realise a maximum
profit? What is the maximum profit per week?
518
MATHEMATICS
Solution Suppose x is the number of pieces of Model A and y is the number of pieces
of Model B
...
Maximise Z = 8000 x + 12000 y
...
e
...
(2)
(Finishing constraint)
...
(4)
The feasible region (shaded) OABC determined by the linear inequalities (2) to (4)
is shown in the Fig 12
...
Note that the feasible region is bounded
...
9
Let us evaluate the objective function Z at each corner point as shown below:
Corner Point
0 (0, 0)
Z = 8000 x + 12000 y
0
A (20, 0)
160000
B (12, 6)
168000 ☎
C (0, 10)
120000
Maximum
We find that maximum value of Z is 1,68,000 at B (12, 6)
...
LINEAR PROGRAMMING
519
EXERCISE 12
...
Reshma wishes to mix two types of food P and Q in such a way that the vitamin
contents of the mixture contain at least 8 units of vitamin A and 11 units of
vitamin B
...
Food P contains
3 units/kg of Vitamin A and 5 units / kg of Vitamin B while food Q contains
4 units/kg of Vitamin A and 2 units/kg of vitamin B
...
2
...
Find the maximum number of cakes which
can be made from 5kg of flour and 1 kg of fat assuming that there is no shortage
of the other ingredients used in making the cakes
...
A factory makes tennis rackets and cricket bats
...
5 hours
of machine time and 3 hours of craftman’s time in its making while a cricket bat
takes 3 hour of machine time and 1 hour of craftman’s time
...
(i) What number of rackets and bats must be made if the factory is to work
at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find
the maximum profit of the factory when it works at full capacity
...
A manufacturer produces nuts and bolts
...
It takes 3 hours on
machine A and 1 hour on machine B to produce a package of bolts
...
50 per package on nuts and Rs 7
...
How
many packages of each should be produced each day so as to maximise his
profit, if he operates his machines for at the most 12 hours a day?
5
...
Each type of screw
requires the use of two machines, an automatic and a hand operated
...
Each machine is available for at the most 4 hours on any day
...
Assuming that he can sell all the screws he manufactures, how many
packages of each type should the factory owner produce in a day in order to
maximise his profit? Determine the maximum profit
...
A cottage industry manufactures pedestal lamps and wooden shades, each
requiring the use of a grinding/cutting machine and a sprayer
...
It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer
to manufacture a shade
...
The profit from
the sale of a lamp is Rs 5 and that from a shade is Rs 3
...
A company manufactures two types of novelty souvenirs made of plywood
...
Souvenirs of type B require 8 minutes each for cutting and 8 minutes
each for assembling
...
The profit is Rs 5 each for type A and Rs 6 each for type
B souvenirs
...
A merchant plans to sell two types of personal computers – a desktop model and
a portable model that will cost Rs 25000 and Rs 40000 respectively
...
Determine
the number of units of each type of computers which the merchant should stock
to get maximum profit if he does not want to invest more than Rs 70 lakhs and if
his profit on the desktop model is Rs 4500 and on portable model is Rs 5000
...
A diet is to contain at least 80 units of vitamin A and 100 units of minerals
...
Food F1 costs Rs 4 per unit food and F2 costs
Rs 6 per unit
...
One unit of food F2 contains 6 units of vitamin A and 3 units of minerals
...
Find the minimum cost for diet
that consists of mixture of these two foods and also meets the minimal nutritional
requirements
...
There are two types of fertilisers F1 and F2
...
After
testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen
and 14 kg of phosphoric acid for her crop
...
What is the minimum cost?
11
...
Let
Z = px + qy, where p, q > 0
...
Each packet (containing 30 g) of food P contains 12 units of calcium, 4 units
of iron, 6 units of cholesterol and 6 units of vitamin A
...
The diet requires atleast 240 units of calcium, atleast 460 units of iron and
at most 300 units of cholesterol
...
Obviously
x ✄ 0, y ✄ 0
...
e
...
(1)
4x + 20y ✄ 460 (constraint on iron), i
...
x + 5y ✄ 115
...
e
...
(3)
x ✄ 0, y ✄ 0
...
The feasible region (shaded) determined by the constraints (1) to (4) is shown in
Fig 12
...
Fig 12
...
Let us evaluate Z at these points:
Corner Point
Z=6x+3y
(2, 72)
228
(15, 20)
150 ☎
(40, 15)
285
Minimum
From the table, we find that Z is minimum at the point (15, 20)
...
The minimum amount
of vitamin A will be 150 units
...
Machines I and II are capable of being operated for
at most 12 hours whereas machine III must be operated for atleast 5 hours a day
...
The number of hours required for producing 1 unit of each of M and N on the three
machines are given in the following table:
Items Number of hours required on machines
M
N
I
1
2
II
2
1
III
1
1
...
How many
of each item should she produce so as to maximise her profit assuming that she can sell
all the items that she produced? What will be the maximum profit?
Solution Let x and y be the number of items M and N respectively
...
(1)
...
(3)
...
ABCDE is the feasible region
(shaded) as shown in Fig 12
...
Observe that
the feasible region is bounded, coordinates of the corner points A, B, C, D and E are
(5, 0) (6, 0), (4, 4), (0, 6) and (0, 4) respectively
...
11
Let us evaluate Z = 600 x + 400 y at these corner points
...
Hence, the
manufacturer has to produce 4 units of each item to get the maximum profit of Rs 4000
...
From these locations, a certain commodity is to be
delivered to each of the three depots situated at A, B and C
...
The cost of
524
MATHEMATICS
transportation per unit is given below:
From/To
Cost (in Rs)
A
B
C
P
160
100
150
Q
100
120
100
How many units should be transported from each factory to each depot in order that
the transportation cost is minimum
...
12):
Let x units and y units of the commodity be transported from the factory at P to
the depots at A and B respectively
...
12
Hence, we have
x ✄ 0, y ✄ 0
and 8 – x – y ✄ 0
i
...
x ✄ 0, y ✄ 0 and x + y ✂ 8
Now, the weekly requirement of the depot at A is 5 units of the commodity
...
Obviously, 5 – x ✄ 0, i
...
x ✂ 5
...
Thus,
i
...
5 – y ✄ 0 , x + y – 4 ✄0
y✂5 , x+y✄ 4
LINEAR PROGRAMMING
525
Total transportation cost Z is given by
Z = 160 x + 100 y + 100 ( 5 – x) + 120 (5 – y) + 100 (x + y – 4) + 150 (8 – x – y)
= 10 (x – 7 y + 190)
Therefore, the problem reduces to
Minimise Z = 10 (x – 7y + 190)
subject to the constraints:
...
(2)
x✂5
...
(4)
y✂5
and
x+y✄4
...
13
(5) is the feasible region (Fig 12
...
Observe that the feasible region is bounded
...
Let us evaluate Z at these points
...
Hence, the optimal transportation strategy will be to deliver 0, 5 and 3 units from
the factory at P and 5, 0 and 1 units from the factory at Q to the depots at A, B and C
respectively
...
e
...
Miscellaneous Exercise on Chapter 12
1
...
How many packets of each food should be used to maximise
the amount of vitamin A in the diet? What is the maximum amount of vitamin A
in the diet?
526
MATHEMATICS
2
...
Brand P, costing Rs 250 per
bag, contains 3 units of nutritional element A, 2
...
Brand Q costing Rs 200 per bag contains 1
...
25 units of element B, and 3 units of element C
...
Determine the number of bags of each brand which should be mixed in order to
produce a mixture having a minimum cost per bag? What is the minimum cost of
the mixture per bag?
3
...
The vitamin contents of one kg food is given below:
Food
Vitamin A Vitamin B Vitamin C
X
1
2
3
Y
2
2
1
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20
...
A manufacturer makes two types of toys A and B
...
If the profit on
each toy of type A is Rs 7
...
5
...
A profit of Rs 1000 is
made on each executive class ticket and a profit of Rs 600 is made on each
economy class ticket
...
However, at least 4 times as many passengers prefer to travel by economy class
than by the executive class
...
What is the maximum profit?
LINEAR PROGRAMMING
527
6
...
They supply to 3 ration shops, D, E and F whose requirements are
60, 50 and 40 quintals respectively
...
50
3
How should the supplies be transported in order that the transportation cost is
minimum? What is the minimum cost?
7
...
The company is to supply oil to three petrol pumps, D, E and F
whose requirements are 4500L, 3000L and 3500L respectively
...
)
From / To
A
B
D
7
3
E
6
4
F
3
2
Assuming that the transportation cost of 10 litres of oil is Re 1 per km, how
should the delivery be scheduled in order that the transportation cost is minimum?
What is the minimum cost?
8
...
The amounts (in kg) of nitrogen, phosphoric acid, potash, and chlorine in a bag of
each brand are given in the table
...
If the grower wants to minimise the amount of nitrogen added to the garden,
how many bags of each brand should be used? What is the minimum amount of
nitrogen added in the garden?
528
MATHEMATICS
kg per bag
Brand P
Brand Q
Nitrogen
3
3
...
5
1
...
Refer to Question 8
...
A toy company manufactures two types of dolls, A and B
...
Further, the production level of dolls of type A can exceed three times the
production of dolls of other type by at most 600 units
...
Variables are sometimes called decision variables and are non-negative
...
Points within and on the boundary of the feasible region represent feasible
solutions of the constraints
...
✄ ✄
LINEAR PROGRAMMING
529
Any point in the feasible region that gives the optimal value (maximum or
minimum) of the objective function is called an optimal solution
...
When Z
has an optimal value (maximum or minimum), where the variables x and y
are subject to constraints described by linear inequalities, this optimal value
must occur at a corner point (vertex) of the feasible region
...
If R is bounded, then the
objective function Z has both a maximum and a minimum value on R and
each of these occurs at a corner point (vertex) of R
...
However, if it exists, it must occur at a corner point of R
...
The method
comprises of the following steps:
(i) Find the feasible region of the linear programming problem and determine
its corner points (vertices)
...
Let M
and m respectively be the largest and smallest values at these points
...
If the feasible region is unbounded, then
(i) M is the maximum value of the objective function, if the open half plane
determined by ax + by > M has no point in common with the feasible
region
...
(ii) m is the minimum value of the objective function, if the open half plane
determined by ax + by < m has no point in common with the feasible
region
...
If two corner points of the feasible region are both optimal solutions of the
same type, i
...
, both produce the same maximum or minimum, then any point
on the line segment joining these two points is also an optimal solution of the
same type
...
The first problem in linear programming was formulated in 1941 by the Russian
mathematician, L
...
L
...
This was the well
known transportation problem
...
Stigler,
described yet another linear programming problem – that of determining an
optimal diet
...
B
...
L
...
C
...
With the advent of computers and the necessary
softwares, it has become possible to apply linear programming model to
increasingly complex problems in many areas
...
In Mathematics alone each generation
builds a new story to the old structure
...
1 Introduction
In our day to day life, we come across many queries such
as – What is your height? How should a football player hit
the ball to give a pass to another player of his team? Observe
that a possible answer to the first query may be 1
...
Such quantities are called scalars
...
Such
quantities are called vectors
...
R
...
and vector quantities like displacement, velocity,
acceleration, force, weight, momentum, electric field intensity etc
...
These two type of
properties, when considered together give a full realisation to the concept of vectors,
and lead to their vital applicability in various areas as mentioned above
...
2 Some Basic Concepts
Let ‘l’ be any straight line in plane or three dimensional space
...
A line with one of these directions prescribed
is called a directed line (Fig 10
...
VECTOR ALGEBRA
425
Fig 10
...
1(iii))
...
Definition 1 A quantity that has magnitude as well as direction is called a vector
...
1(iii)), denoted as AB or
✄✄✄☎
✆
simply as a✂ , and read as ‘vector AB ’ or ‘vector a ’
...
The distance between initial and
terminal points of a vector is called the magnitude (or length) of the vector, denoted as
✄✄✄☎
✆
| AB |, or | a |, or a
...
✆
Note Since the length is never negative, the notation | a | < 0 has no meaning
...
2(i))
...
Then, the vector OP having O and P as its initial and
terminal points, respectively, is called the position vector of the point P with respect
✠✠✠✡
☛
to O
...
, with respect to the origin O
✆
☛ ☛
are denoted by a , b , c , etc
...
2 (ii))
...
2
Direction Cosines
✂✂✂✄
✄
Consider the position vector OP or r ✁ of a point P(x, y, z) as in Fig 10
...
The angles ☎,
✞
✆, ✝ made by the vector r with the positive directions of x, y and z-axes respectively,
are called its direction angles
...
e
...
C
P(x,y,z)
z
r
O
y
B
Y
P
x
A
O
90°
X
A
Fig 10
...
3, one may note that the triangle OAP is right angled, and in it, we
x
☞
have cos ✡ ☛ ✟ r stands for | r |✠
...
Thus, the coordinates of the point P may
r
r
also be expressed as (lr, mr,nr)
...
✆
VECTOR ALGEBRA
Note One may note that l2 + m2 + n2 = 1 but a2 + b2 + c2
10
...
Zero Vector A vector whose initial and terminal points coincide, is called a zero
vector (or null vector), and denoted as 0
...
Or, alternatively otherwise, it may be regarded as
having any direction
...
e
...
The
unit vector in the direction of a given vector a is denoted by aˆ
...
✁ ✁
Collinear Vectors Two or more vectors are said to be collinear if they are parallel to
the same line, irrespective of their magnitudes and directions
...
✝✝✝✞
✝✝✝✞
✂✂✂✄ ✟✠ ✂✂✂✄
Negative of a Vector A vector whose magnitude is the same as that of a given vector
(say, AB ), but direction is opposite to that of it, is called negative of the given vector
...
Remark The vectors defined above are such that any of them may be subject to its
parallel displacement without changing its magnitude and direction
...
Throughout this chapter, we will be dealing with free vectors only
...
Solution The vector OP represents the required
displacement (Fig 10
...
Example 2 Classify the following measures as
scalars and vectors
...
4
428
MATHEMATICS
(iii) 10 Newton
(iv) 30 km/hr
(vi) 20 m/s towards north
Solution
(i) Time-scalar
(iv) Speed-scalar
(ii) Volume-scalar
(v) Density-scalar
Example 3 In Fig 10
...
✁
✁
(ii) Equal vectors : a and c
...
Fig 10
...
1
1
...
2
...
(i) 10 kg
(ii) 2 meters north-west (iii) 40°
(v) 10–19 coulomb
(iv) 40 watt
(vi) 20 m/s2
3
...
(i) time period
(iv) velocity
(ii) distance
(v) work done
(iii) force
4
...
6 (a square), identify the following vectors
...
Answer the following as true or false
...
(ii) Two collinear vectors are always equal in
magnitude
...
6
(iii) Two vectors having same magnitude are collinear
...
VECTOR ALGEBRA
429
10
...
Now consider a situation that a
girl moves from A to B and then from B to C
(Fig 10
...
The net displacement made by the girl from
✂✂✂✄
point A to the point C, is given by the vector AC and
expressed as
✂✂✂✄
✂✂✂✄
Fig 10
...
✆
✄
In general, if we have two vectors a and b (Fig 10
...
8(ii))
...
8
–b
C➆
✝
For example, in Fig 10
...
Then, the
✝
✝
vector a ✞ b , represented by the third side AC of the triangle ABC, gives us the sum
✄
✆
(or resultant) of the vectors a and b i
...
, in triangle ABC (Fig 10
...
8(iii))
...
8 (iii)), i
...
,
☎☎☎
✆
✁✁✁✁
✂
BC =
✄
BC
Then, on applying triangle law from the Fig 10
...
Now, consider a boat in a river going from one bank of the river to the other in a
direction perpendicular to the flow of the river
...
Under the simultaneous influence of these two
velocities, the boat in actual starts travelling with a different velocity
...
e
...
✆
✆
If we have two vectors a and b represented
by the two adjacent sides of a parallelogram in
magnitude and direction (Fig 10
...
This is known as
the parallelogram law of vector addition
...
9
Note From Fig 10
...
Thus, we may say that the two laws of vector
addition are equivalent to each other
...
10)
...
10, we have,
✆✆✆✝
✆✆✆✝
✝
and
AD = BC = b
✝
DC = AB = a
...
10
✂ ✂
AC = AD + DC = b + a
Hence
☎
☎
☎
☎
a✞b = b ✞a
✂ ✂
✂
Property 2 For any three vectors a , b and c
✂
✂
✂
✂
✂ ✂
( a ✟ b ) ✟ c = a ✟ (b ✟ c )
✂ ✂
(Associative property)
✁✁✁✂ ✁✁✁✂
✂
✁✁✁✂
Proof Let the vectors a, b and c be represented by PQ, QR and RS , respectively,
as shown in Fig 10
...
☎
☎
Fig 10
...
✠
Note that for any vector a , we have
☛
☛
☛
☛
☛
a✡0 = 0✡ a ☞ a
☛
Here, the zero vector 0 is called the additive identity for the vector addition
...
5 Multiplication of a Vector by a Scalar
✌
✌
Let a be a given vector and ✍ a scalar
...
Note that,
✎
✌
✌
✍ a is also a vector, collinear to the vector a
...
✌
✌
Also, the magnitude of vector ✍ a is | ✍ | times the magnitude of the vector a , i
...
,
✑
✑
| ✏a | = | ✏ | | a |
A geometric visualisation of multiplication of a vector by a scalar is given
in Fig 10
...
✆
✷
❛
✶
➊
✁
✷
☎
✄
✂
Fig 10
...
The vector – a is
✎
called the negative (or additive inverse) of vector a and we always have
✕
✕
✞
✞
✞
a ✖ (– a ) = (– a ) ✝ a ✗ 0
Also, if
✘
=
✛
✛
1
✙ , provided a ✚ 0, i
...
a is not a null vector, then
|a |
✕
✕
| ✒a |✓| ✒ | | a | =
1 ✙
✙ | a |✜ 1
|a |
VECTOR ALGEBRA
433
So, ✝ a represents the unit vector in the direction of a
...
✂
10
...
1 Components of a vector
Let us take the points A(1, 0, 0), B(0, 1, 0) and C(0, 0, 1) on the x-axis, y-axis and
z-axis, respectively
...
13)
...
13
Now, consider the position vector OP of a point P (x, y, z) as in Fig 10
...
Let P1
be the foot of the perpendicular from P on the plane XOY
...
14
parallel to z-axis
...
✆✆✆✄
✆✆✆✄
✆✆✆✄
Similarly, QP1 ☎ OS ☎ yjˆ and OQ ☎ xiˆ
...
Here, x, y and z are called
✁
as the scalar components of r , and xiˆ, yjˆ and zkˆ are called the vector components
✁
of r along the respective axes
...
✁
The length of any vector r ✂ xiˆ ✄ yjˆ ✄ zkˆ , is readily determined by applying the
Pythagoras theorem twice
...
14)
✁
| OP 1 | =
✟✟✟✟✠
✟✟✟✠
| OQ |2 +|QP1|2 ✝ x 2 ✞ y 2 ,
and in the right angle triangle OP1P, we have
✁
| OP 1 | =
☞☞☞✌
☞☞☞✌
| OP1 |2 ✡ | P1P |2 ☛ ( x 2 ✡ y 2 ) ✡ z 2
✁
Hence, the length of any vector r ✂ xiˆ ✄ yjˆ + zkˆ is given by
✍
| r | = | xiˆ ✞ yjˆ ✞ zkˆ | = x 2 ✞ y 2 ✞ z 2
✆
✆
If a and b are any two vectors given in the component form a1iˆ ✄ a2 ˆj + a3kˆ and
b1iˆ ✄ b2 ˆj ✄ b3 kˆ , respectively, then
✎
✎
(i) the sum (or resultant) of the vectors a and b is given by
✆
✆
a ✏ b = (a1 ✄ b1 )iˆ ✄ (a2 ✄ b2 ) ˆj ✄ (a3 ✄ b3 )kˆ
✆
✆
(ii) the difference of the vector a and b is given by
✆
✆
a ✑ b = ( a1 ✒ b1 )iˆ ✄ ( a2 ✒ b2 ) ˆj ✄ (a3 ✒ b3 )kˆ
✎
✎
(iii) the vectors a and b are equal if and only if
a1 = b1, a2 = b2 and a3 = b3
✓
(iv) the multiplication of vector a by any scalar ✔ is given by
✓
✕a
= ( ✖a1 )iˆ ✄ (✖a2 ) ˆj ✄ (✖a3 )kˆ
VECTOR ALGEBRA
435
The addition of vectors and the multiplication of a vector by a scalar together give
the following distributive laws:
Let a and b be any two vectors, and k and m be any scalars
...
In fact, two vectors a and b are collinear if and only
if there exists a nonzero scalar ✠ such that b ☞ ✌a
...
e
...
(iii) In case if it is given that l, m, n are direction cosines of a vector, then liˆ ✝ mjˆ ✝ nkˆ
= (cos ✓ )iˆ ✝ (cos ✔) ˆj ✝ (cos ✕ )kˆ is the unit vector in the direction of that vector,
where ✖, ✗ and ✘ are the angles which the vector makes with x, y and z axes
respectively
...
Solution Note that two vectors are equal if and only if their corresponding components
✙
✙
are equal
...
Is | a | | b | ? Are the vectors a and b
Example 5 Let a
equal?
✝
✝
2
2
Solution We have | a | ☎ 12 ✆ 22 ☎ 5 and | b | ☎ 2 ✆ 1 ☎ 5
✂
✂
So, | a | | b |
...
✂
Example 6 Find unit vector in the direction of vector a 2iˆ ✁ 3 ˆj ✁ kˆ
1 ✠
✞
Solution The unit vector in the direction of a vector a is given by aˆ ✟ ✠ a
...
Solution The unit vector in the direction of the given vector a✎ is
1 ✠
1 ˆ
1 ˆ 2 ˆ
aˆ ✟ ✠ a =
(i ✏ 2 ˆj ) ✟
i✏
j
|a|
5
5
5
✞
Therefore, the vector having magnitude equal to 7 and in the direction of a is
✑
7 ˆ 14 ˆ
✓ 1 ✒ 2 ✒✔
i✚
j
i✕
j✗ =
5
5
5 ✙
✘ 5
7 a = 7✖
Example 8 Find the unit vector in the direction of the sum of the vectors,
✂
✂
a 2iˆ ✁ 2 ˆj – 5kˆ and b 2iˆ ✁ ˆj ✁ 3kˆ
...
✞
Solution Note that the direction ratio’s a, b, c of a vector r ☎ xiˆ ✆ yjˆ ✆ zkˆ are just
the respective components x, y and z of the vector
...
Further, if l, m and n are the direction cosines of the given
vector, then
1
a
,
l✠ ✡ ✠
|r |
6
1
b
c
✟2
✡
, n✠ ✡ ✠
as |r | ✠ 6
m✠ ✡ ✠
|r |
|r |
6
6
2 ☞
☛ 1 1
,
,–
✍
...
5
...
15)
...
Using the properties of vector addition, the
above equation becomes
✑✑✑✑✞ ✑✑✑✞
✑✑✑✑✞
P1P2 = OP2 ✝ OP1
i
...
✑✑✑✑✞
P1P2 = ( x2iˆ ✆ y2 ˆj ✆ z2 kˆ ) ✝ ( x1iˆ ✆ y1 ˆj ✆ z1kˆ )
= ( x2 ✝ x1 )iˆ ✆ ( y2 ✝ y1 ) ˆj ✆ ( z2 ✝ z1 ) kˆ
✑✑✑✑✞
The magnitude of vector P1P2 is given by
✑✑✑✑✞
P1P2 =
( x2 ✒ x1 ) 2 ✓ ( y2 ✒ y1 )2 ✓ ( z2 ✒ z1 ) 2
Fig 10
...
Solution Since the vector is to be directed from P to Q, clearly P is the initial point
✁
and Q is the terminal point
...
i
...
10
...
3 Section formula
✁
✁
Let P and Q be two points represented by the position vectors OP and OQ , respectively,
with respect to the origin O
...
16)
and externally (Fig 10
...
Here, we intend to find
☎☎☎✆
the position vector OR for the point R with respect
to the origin O
...
Case I When R divides PQ internally (Fig 10
...
✁
✁
Fig 10
...
Now from triangles ORQ and OPR, we have
✁
✁
✁
✁
✁
✞✞✞✟
☎☎☎✆
☎☎☎✆
✆
✆
RQ = OQ ✂ OR ✝ b ✂ r
and
Therefore, we have
PR = OR ✠ OP ✡ r ✠ a ,
✁
✁
☞
✍
or
☞
m (b ✂ r ) = n (r ☛ a )
✁
r =
(Why?)
✍
mb ✌ na
m✌n
(on simplification)
Hence, the position vector of the point R which divides P and Q internally in the
ratio of m : n is given by
☎☎☎✆
✍
✍
mb ✌ na
OR = m n
✌
VECTOR ALGEBRA
439
Case II When R divides PQ externally (Fig 10
...
We leave it to the reader as an exercise to verify
that the position vector of the point R which divides
the line segment PQ externally in the ratio
PR
QR
m : n ✄ i
...
✆
✂
m✁
is given by
n ☎✝
✡
✡
mb ✠ na
Fig 10
...
And therefore, from Case I, the
✞✞✞✟
☛☛☛☞
midpoint R of PQ , will have its position vector as
✡
✡
a ✌b
OR = 2
✞✞✞✟
✞✞✞✟
✟
✟
Example 11 Consider two points P and Q with position vectors OP ✍ 3a ✎ 2b and
☛☛☛☞
☞
☞
OQ ✏ a ✑ b
...
Solution
(i) The position vector of the point R dividing the join of P and Q internally in the
ratio 2:1 is
✡
✡
✡
✡
✡
✡
✡
✡
✡
2( a ✌ b ) ✌ (3a ✠ 2b ) 5a
✒
OR =
2 ✌1
3
(ii) The position vector of the point R dividing the join of P and Q externally in the
ratio 2:1 is
✞✞✞✟
✡ ✡
2( a ✌ b ) ✠ (3a ✠ 2b )
✒ 4b ✠ a
OR =
2 ✠1
✓✓✓✔
Example 12 Show that the points A(2iˆ ✕ ˆj ✑ kˆ ), B(iˆ ✕ 3 ˆj ✕ 5kˆ), C(3iˆ ✕ 4 j ✕ 4kˆ) are
the vertices of a right angled triangle
...
EXERCISE 10
...
Compute the magnitude of the following vectors:
✞
1 ˆ 1 ˆ 1 ˆ
✞
b ☎ 2iˆ ✝ 7 ˆj ✝ 3kˆ; c ☎
i✆
j✝
k
3
3
3
2
...
3
...
✞
a ☎ iˆ ✆ ˆj ✆ k;
4
...
5
...
✁
✁
✁
6
...
✁
7
...
✁
8
...
✁
✁
9
...
10
...
11
...
12
...
13
...
14
...
15
...
Find the position vector of the mid point of the vector joining the points P(2, 3, 4)
and Q(4, 1, –2)
...
Show that the points A, B and C with position vectors, a
✂
✂
3iˆ ✁ 4 ˆj ✁ 4kˆ,
b 2iˆ ✁ ˆj ✄ kˆ and c iˆ ✁ 3 ˆj ✁ 5kˆ , respectively form the vertices of a right angled
triangle
...
In triangle ABC (Fig 10
...
18
(D) AB ✡ CB ☛ CA ☞ 0
✂
✂
19
...
10
...
An other algebraic
operation which we intend to discuss regarding vectors is their product
...
But in case of functions, we may multiply them in two ways, namely,
multiplication of two functions pointwise and composition of two functions
...
Based upon these two types of products for vectors, they have
found various applications in geometry, mechanics and engineering
...
10
...
1 Scalar (or dot) product of two vectors
✂
✂
✂ ✂
Definition 2 The scalar product of two nonzero vectors a and b , denoted by a ✓ b , is
442
MATHEMATICS
✁
✁
defined as
✁
✁
a b = | a | | b | cos ✂ ,
✁
✁
where, ✟ is the angle between a and b , 0 ✄ ✂ ✄ ☎ (Fig 10
...
✁
✁
✁
✁
If either a ✆ 0 or b ✆ 0, then ✟ is not defined, and in this case,
✠
Fig 10
...
✁
✁
a b is a real number
...
Let a and b be two nonzero vectors, then a b ✆ 0 if and only if a and b are
perpendicular to each other
...
e
...
If ✟ = 0, then a b ✆ | a | | b |
✁
✁
✁
In particular, a a ✆ | a |2 , as ✟ in this case is 0
...
If ✟ = ☞, then a b ✆ ✌ | a | | b |
✁
✁
✁
In particular, a (✌ a ) ✆ ✌ | a |2 , as ✟ in this case is ☞
...
In view of the Observations 2 and 3, for mutually perpendicular unit vectors
iˆ, ˆj and kˆ, we have
iˆ iˆ ✆ ˆj ˆj = kˆ ✍ kˆ ✎ 1,
iˆ ˆj ✆ ˆj kˆ = kˆ ✏ iˆ ✑ 0
✒
✒
6
...
b ✘
✁ ✚
✆ cos
✙ ✁
✛ | a || b | ✜
–1 ✗
7
...
i
...
✁
✁
✢
✢
a b = b ✏a
(Why?)
Two important properties of scalar product
✣
✣
✣
Property 1 (Distributivity of scalar product over addition) Let a, b and c be
any three vectors, then
✁
✠
✠
✁
✁
a (b ✤ c ) = a ✝ b
✥
✠
✠
a ✝c
VECTOR ALGEBRA
443
☞
☞
Property 2 Let a and b be any two vectors, and ✌ be any scalar
...
6
...
20)
...
The vector p
❇
❇
☎
✆
❧
❆
❧
❈
♣
✵
✭✞ ❁
❈
✵
❁ ✝✞ ✁
♣
✵
✭✝✞ ❁
✭ ✁
✭
♣
❈
♣
❧
✭
✵
❁ ✠✡✞ ✁
✁
✁
❈
❛
✆
✵
✭✶✟✞ ❁
✵
❁ ✶✟✞ ✁
✂
❆
❇
❆
✵
✭✠✡✞ ❁
Fig 10
...
For example, in each of the following figures (Fig 10
...
Observations
✞
1
...
✠
✞
2
...
If ✔ = 0, then the projection vector of AB will be AB itself and if ✔ = ✗, then the
✘✘✘✙
✘✘✘✙
projection vector of AB will be BA
...
2
2
✓
Remark If ✜, ✢ and ✣ are the direction angles of vector a ✤ a1iˆ ✥ a2 ˆj ✥ a3 kˆ , then its
direction cosines may be given as
4
...
i
...
, the scalar components a1, a2 and a3 of the vector a ,
✯
are precisely the projections of a along x-axis, y-axis and z-axis, respectively
...
✓ ✓
✓
✓
Solution Given a ✍ b ✤ 1, | a | ✤ 1 and | b | ✤ 2
...
☎
Solution The angle ✟ between two vectors a and b is given by
✝ ✝
a✆b
cos✟ = ✝ ✝
| a || b |
✠
Now
✠
a ✞ b = (iˆ ✁ ˆj ✂ kˆ) ✡ ( iˆ ✂ ˆj ✁ kˆ ) 1 ✂ 1 ✂ 1 ✂ 1
...
hence the required angle is
Solution We know that two nonzero vectors are perpendicular if their scalar product
is zero
...
✄
✄
✄ ✄
☎ ☎
☎ ☎
4iˆ ✂ 4 ˆj ✁ 2kˆ
(6iˆ ✁ 2 ˆj ✂ 8kˆ ) ✡ (4iˆ ✂ 4 ˆj ✁ 2kˆ) 24 ✂ 8 ✂ 16 0
...
2iˆ ✁ 3 ˆj ✁ 2kˆ on the vector
☎
✖
Solution The projection of vector a on the vector b is given by
(2 ✙ 1 ✚ 3 ✙ 2 ✚ 2 ✙ 1) 10 5
1 ✘ ✘
✛
✛ 6
✘ (a ✗ b ) =
6 3
|b |
(1) 2 ✚ (2) 2 ✚ (1)2
✄
✄
✄
✄
✄
✄
Example 17 Find | a ✂ b | , if two vectors a and b are such that | a | 2, | b | 3
☎ ☎
and a ✜ b ✢ 4
...
a ✣ a ✞ b ✣ b ✞ a ✤ b ✞ b
446
MATHEMATICS
✄ ✄
✄
✄
= | a |2 2( a ✁ b )✂ | b |2
= (2)2
✄ ✄
Therefore
|a b | =
2(4) ✂ (3) 2
5
✠ ✠
☎
✠ ✠
✠
Example 18 If a is a unit vector and ( x ✆ a ) ✝ ( x ✞ a ) ✟ 8 , then find | x |
...
Also,
✎ ✎ ✎ ✎
( x ☞ a ) ✌ ( x ✍ a) = 8
✓ ✓ ✓ ✓ ✓ ✓ ✓ ✓
x✏ x ✑ x✏a ✒ a✏ x ✒ a✏a = 8
or
✄
✓
| x |2 1 = 8 i
...
| x | 2 = 9
or
✎
| x | = 3 (as magnitude of a vector is non negative)
...
✖ ✖
✖ ✖
Solution The inequality holds trivially when either a ✕ 0 or b ✕ 0
...
So, let us assume that | a✄ | ✘ 0 ✘ | b |
...
Solution The inequality holds trivially in case either
b
A
a
B
✄ ✄ ✄
✄ ✄ ✄ ✄
a ✗ 0 or b ✗ 0 (How?)
...
Then,
Fig 10
...
e
...
Example 21 Show that the points A (✄2iˆ 3 ˆj 5kˆ ), B( iˆ 2 ˆj 3kˆ) and C(7iˆ ✄ kˆ)
are collinear
...
✞✞✞✟ ✞✞✞✟ ✞✞✞✟ ✟
✠Note In Example 21, one may note that although AB ☛ BC ☛ CA ☞ 0 but the
points A, B and C do not form the vertices of a triangle
...
3
✟
✟
1
...
3 and 2 ,
2
...
Find the projection of the vector iˆ ✄ ˆj on the vector iˆ ˆj
...
Find the projection of the vector iˆ 3 ˆj 7kˆ on the vector 7iˆ ✄ ˆj 8kˆ
...
Show that each of the given three vectors is a unit vector:
1 ˆ
1
1 ˆ
(2i ✒ 3 ˆj ✒ 6kˆ ), (3iˆ ✓ 6 ˆj ✒ 2kˆ ),
(6i ✒ 2 ˆj ✓ 3kˆ)
7
7
7
Also, show that they are mutually perpendicular to each other
...
Find | a | and | b | , if (a ✁ b ) ✂ (a ✄ b ) ☎ 8 and | a |☎ 8 | b |
...
Evaluate the product (3a ✄ 5b ) ✂ (2a ✁ 7b )
...
Find the magnitude of two vectors a and b , having the same magnitude and
such that the angle between them is 60o and their scalar product is
✝
✞
✝
✝
✝
1
...
Find | x | , if for a unit vector a , ( x ✟ a ) ✠ ( x ✡ a ) ☛ 12
...
If a ☎ 2iˆ ✁ 2 ˆj ✁ 3kˆ, b ☎ ✄ iˆ ✁ 2 ˆj ✁ kˆ and c ☎ 3iˆ ✁ ˆj are such that a✍ ☞ ✌ b is
✎
perpendicular to c , then find the value of ✏
...
Show that | a | b ✁ | b | a is perpendicular to | a | b ✄ | b | a , for any two nonzero
✍
✍
vectors a and b
...
If a ✑ a ✒ 0 and a ✑ b ✒ 0 , then what can be concluded about the vector b ?
13
...
14
...
But the converse need not be
true
...
15
...
[✓ABC is the angle between the vectors BA
✖✖✖✆
and BC ]
...
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, –1) are collinear
...
Show that the vectors 2iˆ ✄ ˆj ✁ kˆ, iˆ ✄ 3 ˆj ✄ 5kˆ and 3iˆ ✄ 4 ˆj ✄ 4kˆ form the vertices
of a right angled triangle
...
If a is a nonzero vector of magnitude ‘a’ and ✏ a nonzero scalar, then ✏ a is unit
vector if
(B) ✏ = – 1
(C) a = | ✏ |
(D) a = 1/| ✏ |
(A) ✏ = 1
10
...
3 Vector (or cross) product of two vectors
In Section 10
...
In this system, when the positive x-axis is rotated counterclockwise
VECTOR ALGEBRA
449
into the positive y-axis, a right handed (standard) screw would advance in the direction
of the positive z-axis (Fig 10
...
In a right handed coordinate system, the thumb of the right hand points in the
direction of the positive z-axis when the fingers are curled in the direction away from
the positive x-axis toward the positive y-axis (Fig 10
...
Fig 10
...
23)
...
e
...
23
direction of nˆ
...
Observations
✠
✠
1
...
✠
✠
✠
✠
✠
2
...
Then a ☛ b ☞ 0 if and only if a and b
are parallel (or collinear) to each other, i
...
,
✠
✠
a☛b = 0✌ a ✍b
450
MATHEMATICS
✂ ✂ ✂
✝
✝
✝
In particular, a a ✁ 0 and a ✄ (☎ a ) ✆ 0 , since in the first situation, ✟ = 0
and in the second one, ✟ ✞ ✠ , making the value of sin ✟ to be 0
...
If ☛ ☞
✡
✝
✝
✝ ✝
then a ✄ b ✆ | a || b |
...
In view of the Observations 2 and 3, for mutually perpendicular
unit vectors iˆ, ˆj and kˆ (Fig 10
...
24
✍
✍
5
...
It is always true that the vector product is not commutative, as a ✌ b = ☎ b ✄ a
...
e
...
25 (i)
...
e
...
25(ii)
...
25 (i), (ii)
✝
Thus, if we assume a and b to lie in the plane of the paper, then nˆ and nˆ1 both
will be perpendicular to the plane of the paper
...
i
...
nˆ1 ✓ ✔ nˆ
...
In view of the Observations 4 and 6, we have
ˆj iˆ ☎ ✄ kˆ, kˆ ˆj ☎ ✄ iˆ and iˆ kˆ ☎ ✄ ˆj
...
If a and b represent the adjacent sides of a triangle then its area is given as
1 ✝ ✝
|a ✆b |
...
26,
Area of triangle ABC =
✁
1
AB ✞ CD
...
26
✟
But AB ☎ | b | (as given), and CD = | a | sin ✠
...
2
2
9
...
From Fig 10
...
DE
...
Thus,
✁ ✁
Fig 10
...
We now state two important properties of vector product
...
Then their cross product may be given by
1
2
3
iˆ
ˆj
kˆ
a ✂ b = a1
b1
a2
b2
a3
b3
✄
✄
Explanation We have
a ☎ b = (a1iˆ ✁ a2 ˆj ✁ a3kˆ ) ☎ (b1iˆ ✁ b2 ˆj ✁ b3kˆ )
= a1b1 (iˆ ☎ iˆ) ✁ a1b2 (iˆ ☎ ˆj ) ✁ a1b3 (iˆ ☎ kˆ ) ✁ a2b1 ( ˆj ☎ iˆ)
+ a2b2 ( ˆj ☎ ˆj ) ✁ a2b3 ( ˆj ☎ kˆ)
+ a3b1 ( kˆ ☎ iˆ) ✁ a3b2 (kˆ ☎ ˆj ) ✁ a3b3 ( kˆ ☎ kˆ)
(by Property 1)
= a1b2 (iˆ ☎ ˆj ) ✆ a1b3 (kˆ ☎ iˆ) ✆ a2b1 (iˆ ☎ ˆj )
+ a2b3 ( ˆj ☎ kˆ ) ✁ a3b1 (kˆ ☎ iˆ) ✆ a3b2 ( ˆj ☎ kˆ)
(as iˆ ☎ iˆ ✝ ˆj ☎ ˆj ✝ kˆ ☎ kˆ ✝ 0 and iˆ ☎ kˆ ✝ ✆ kˆ ☎ iˆ, ˆj ☎ iˆ ✝ ✆ iˆ ☎ ˆj and kˆ ☎ ˆj ✝ ✆ ˆj ☎ kˆ)
= a1b2 kˆ ✆ a1b3 ˆj ✆ a2b1kˆ ✁ a2b3iˆ ✁ a3b1 ˆj ✆ a3b2iˆ
(as iˆ ☎ ˆj ✝ kˆ, ˆj ☎ kˆ ✝ iˆ and kˆ ☎ iˆ ✝ ˆj )
= (a2b3 ✆ a3b2 )iˆ ✆ (a1b3 ✆ a3b1 ) ˆj ✁ ( a1b2 ✆ a2b1 ) kˆ
ˆj
iˆ
= a1 a2
b1 b2
kˆ
a3
b3
Example 22 Find | a ☎ b |, if a ✝ 2iˆ ✁ ˆj ✁ 3kˆ and b ✝ 3iˆ ✁ 5 ˆj ✆ 2kˆ
Solution We have
✄
iˆ
✄
ˆj
kˆ
a✂ b = 2 1 3
3 5 ✞2
= iˆ(✆2 ✆ 15) ✆ (✆ 4 ✆ 9) ˆj ✁ (10 – 3)kˆ ✝ ✆17iˆ ✁ 13 ˆj ✁ 7kˆ
Hence
| a ☎b | =
( ✟17) 2 ✠ (13) 2 ✠ (7)2 ✡ 507
VECTOR ALGEBRA
453
✁ ✁
Example 23 Find a unit vector perpendicular to each of the vectors (a b ) and
✁ ✁
(a ✂ b ), where a✁ ✄ iˆ
✁
ˆj kˆ, b ✄ iˆ 2 ˆj 3kˆ
...
Thus, another unit
✏
✏
1 ˆ 2 ˆ 1 ˆ
i☛
j☞
k
...
6
6
Example 24 Find the area of a triangle having the points A(1, 1, 1), B(1, 2, 3)
and C(2, 3, 1) as its vertices
...
The area of the given triangle
is
1 ✓✓✓✔ ✓✓✓✔
| AB ✒ AC |
...
iˆ
ˆj
✝ ✝
a✆b = 3
kˆ
1 4 ✞ 5iˆ ✟ ˆj ✠ 4kˆ
1 ✠1 1
Now
✄ ✄
| a☎b | =
Therefore
and hence, the required area is
25 ✡ 1 ✡ 16 ☛ 42
42
...
4
✄
✄ ✄
✄
1
...
✄ ✄
✄ ✄
2
...
☞
3
...
4
...
Find ✏ and ➭ if (2iˆ ✁ 6 ˆj ✁ 27kˆ ) ☎ (iˆ ✁ ✑ˆj ✁ ✒kˆ) 0
...
Given that a ✓ b ✔ 0 and a ✆ b ✔ 0
...
Let the vectors a, b , c be given as a1iˆ ✁ a2 ˆj ✁ a3kˆ, b1iˆ ✁ b2 ˆj ✁ b3kˆ,
✄ ✄ ✄ ✄ ✄ ✄ ✄
c1iˆ ✁ c2 ˆj ✁ c3 kˆ
...
✕ ✕
✕ ✕
✝ ✝ ✝
8
...
Is the converse true? Justify your
answer with an example
...
Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)
...
Find the area of the parallelogram whose adjacent sides are determined by the
✄
✄
vectors a iˆ ✁ ˆj ✂ 3kˆ and b 2iˆ ✁ 7 ˆj ✂ kˆ
...
Let the vectors a and b be such that | a |✆ 3 and | b |✆
☎
☎
2
✄ ✄
, then a ✞ b is a
3
unit vector, if the angle between a and b is
(B) ✠/4
(C) ✠/3
(D) ✠/2
(A) ✠/6
12
...
✎
☞
☞
Solution Let r ✌ x i ✍ y j be a unit vector in XY-plane (Fig 10
...
Then, from the
figure, we have x = cos ✏ and y = sin ✏ (since | r✑ | = 1)
...
(1)
Clearly,
✘
|r | =
cos 2 ✙ ✚ sin 2 ✙ ✛ 1
Fig 10
...
28) traces the circle x2 + y2 = 1
counterclockwise, and this covers all possible directions
...
456
MATHEMATICS
ˆj kˆ, 2iˆ 5 ˆj, 3iˆ 2 ˆj ✁ 3kˆ and iˆ ✁ 6 ˆj ✁ kˆ are the position
Example 27 If iˆ
✂✂✂✄
vectors of points A, B, C and D respectively, then find the angle between AB and
☎☎☎✆
✂✂✂✄
✝✝✝✞
CD
...
Solution Note that if ✟ is the angle between AB and CD, then ✟ is also the angle
✠✠✠✡
✠✠✠✡
between AB and CD
...
This shows that AB and CD are collinear
...
2
✎
✎
✎
✎
Example 28 Let a , b and c be three vectors such that | a |✌ 3, | b |✌ 4, | c |✌ 5 and
✧ ✧ ✧
each one of them being perpendicular to the sum of the other two, find | a ✦ b ✦ c |
...
✎ ✎
✎
c |2 = (a b
✎ ✎
= a★a
✎
✎
✎
✎
= | a |2 | b |2 | c |2
= 9 + 16 + 25 = 50
Therefore
✎ ✎ ✎ =
c|
|a b
✎
✎ ✎
c ) ★ (a b
✎ ✎ ✎ ✎ ✎ ✎ ✎ ✎
a ★ (b c ) b ★ b b ★ (a c )
✎ ✎ ✎✎
+ c
...
c
✎
✎ ✎
c ) 2 ✌ (a b
50 ✩ 5 2
c)
VECTOR ALGEBRA
✄
✄
✄
457
✄
Example 29 Three vectors a, b and c satisfy the condition a ✁ b ✁ c ✂ 0
...
✄
✄
✄
✄
✟
✟
Solution Since a ✁ b ✁ c ✂ 0 , we have
✟
✟
a ✝ (a ✞ b ✞ c ) = 0
✟ ✟
✟ ✟
✟ ✟
✟ ✟
✟ ✟
a ✝ a ✞ a ✝b ✞ a ✝c = 0
or
☛
2
✓
2
a ✝ b ✞ a ✝c = ✠ a
Therefore
✏
✡ ✠1
...
✒ ✑16
...
(3)
Adding (1), (2) and (3), we have
✟ ✟
✟ ✟
✟ ✟
2 ( a ✝ b ✞ b ✝ c ✞ a ✝ c ) = – 21
2➭ = – 21, i
...
, ➭ =
or
✕21
2
Example 30 If with reference to the right handed system of mutually perpendicular
✟
✟
✟
unit vectors iˆ, ˆj and kˆ, ✖ ✆ 3iˆ ✗ ˆj, ✘ ✆ 2iˆ ✞ ˆj – 3kˆ , then express ✘ in the form
✟ ✟
✟
✟
✟
✚
✟
✘ ✆ ✘1 ✞ ✘ 2 , where ✘1 is parallel to ✖ and ✘ 2 is perpendicular to ✙
...
e
...
✟
✟ ✟
✘ 2 ✆ ✘ ✗ ✘1 = (2 ✗ 3✛ )iˆ ✞ (1 ✞ ✛ ) ˆj ✗ 3kˆ
...
i
...
,
Now
3(2 ✣ 3✤ ) ✣ (1 ✥ ✤ ) = 0
or
Therefore
✦=
1
2
✪
✟
1
3
3
1
✘1 = iˆ ✕ ˆj and ✧2 ★ iˆ ✩ ˆj – 3kˆ
2
2
2
2
458
MATHEMATICS
Miscellaneous Exercise on Chapter 10
1
...
2
...
3
...
Determine the girl’s displacement from her initial point of
departure
...
If a b ✁ c , then is it true that | a✆ |✄| b | ☎ | c✆ | ? Justify your answer
...
Find the value of x for which x(iˆ ☎ ˆj ☎ kˆ) is a unit vector
...
Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
✆
✆
a ✄ 2iˆ ☎ 3 ˆj ✝ kˆ and b ✄ iˆ ✝ 2 ˆj ☎ kˆ
...
If a ✄ iˆ ☎ ˆj ☎ kˆ , b ✄ 2iˆ ✝ ˆj ☎ 3kˆ and c ✄ iˆ ✝ 2 ˆj ☎ kˆ , find a unit vector parallel
✂
✂
✂
to the vector 2a – b ✁ 3c
...
Show that the points A (1, – 2, – 8), B (5, 0, – 2) and C (11, 3, 7) are collinear, and
find the ratio in which B divides AC
...
Find the position vector of a point R which divides the line joining two points
✆
✆
✆
✆
P and Q whose position vectors are (2a ☎ b ) and (a – 3b ) externally in the ratio
1 : 2
...
10
...
Find the unit vector parallel to its diagonal
...
11
...
3 3 3
✂
✆
✆
✆
12
...
Find a vector d
✆
✆
✠ ✠
which is perpendicular to both a and b , and c ✞ d ✟ 15
...
The scalar product of the vector iˆ ☎ ˆj ☎ kˆ with a unit vector along the sum of
vectors 2iˆ ☎ 4 ˆj ✝ 5kˆ and ✡iˆ ☎ 2 ˆj ☎ 3kˆ is equal to one
...
✆ ✆ ✆
14
...
VECTOR ALGEBRA
✄ ✄
✄ ✄
✄
15
...
Choose the correct answer in Exercises 16 to 19
...
If ✟ is the angle between two vectors a and b , then a ✝ b ✞ 0 only when
(A) 0 ✡ ☛ ✡
✠
(B) 0 ☞ ☛ ☞
2
(C) 0 < ✟ < ✌
2
(D) 0 ✍ ✟ ✍ ✌
✆
✆
✠
✏
✏
17
...
Then a ✎ b
is a unit vector if
(A) ☛ ✑
✠
4
(B) ☛ ✑
✠
(C) ☛ ✑
3
✠
(D) ☛ ✑
2
18
...
( ˆj ✒ kˆ) ˆj ✁ (iˆ ✒ kˆ) kˆ ✁ (iˆ ✒ ˆj ) is
(A) 0
(B) –1
(C) 1
✆
✆
(D) 3
2✠
3
✄ ✄
✄ ✄
19
...
✓ The scalar components of a vector are its direction ratios, and represent its
projections along the respective axes
...
460
MATHEMATICS
The vector sum of two coinitial vectors is given by the diagonal of the
parallelogram whose adjacent sides are the given vectors
...
✝
✝
✝
✄
a
✁
For a given vector a , the vector aˆ ✂ ✄ gives the unit vector in the direction
|a |
☎
of a
...
(i) internally, is given by
m✞n
✄ ✄
mb ✟ na
(ii) externally, is given by
...
✆ ✆
✍✍
Also, when a ✌ b is given, the angle ‘✠’ between the vectors a and b may be
determined by
✏✏
a ✎b
cos ✠ = ✏ ✏
| a || b |
✆ ✆
If ✠ is the angle between two vectors a and b , then their cross product is
given as
✍ ✍ ✆ ✆
a ✑ b = | a || b | sin ☞ nˆ
✆ ✆
where nˆ is a unit vector perpendicular to the plane containing a and b
...
✆ ✆
If we have two vectors a and b , given in component form as
✆
✆
a ☛ a iˆ ✒ a ˆj ✒ a kˆ and b ☛ b iˆ ✒ b ˆj ✒ b kˆ and ✝ any scalar,
1
2
3
1
2
3
VECTOR ALGEBRA
then
461
✁ ✁
a b = (a1 ✂ b1 ) iˆ ✂ (a2 ✂ b2 ) ˆj ✂ ( a3 ✂ b3 ) kˆ ;
☎
✄a = (✆a1 )iˆ ✂ (✆a2 ) ˆj ✂ (✆a3 ) kˆ ;
✝✝
a
...
c2
Historical Note
The word vector has been derived from a Latin word vectus, which means
“to carry”
...
William Rowen Hamilton
(1805-1865) an Irish mathematician was the first to use the term vector for a
directed line segment in his book Lectures on Quaternions (1853)
...
Though, we must
mention here that in practice, the idea of vector concept and their addition was
known much earlier ever since the time of Aristotle (384-322 B
...
), a Greek
philosopher, and pupil of Plato (427-348 B
...
That time it was supposed to be
known that the combined action of two or more forces could be seen by adding
them according to parallelogram law
...
In 1586 A
...
, he analysed the principle of
geometric addition of forces in his treatise DeBeghinselen der Weeghconst
(“Principles of the Art of Weighing”), which caused a major breakthrough in the
development of mechanics
...
In the 1880, Josaih Willard Gibbs (1839-1903), an American physicist
and mathematician, and Oliver Heaviside (1850-1925), an English engineer, created
what we now know as vector analysis, essentially by separating the real (scalar)
462
MATHEMATICS
part of quaternion from its imaginary (vector) part
...
This book gave a systematic
and concise account of vectors
...
Heaviside and P
...
Tait (1831-1901)
who contributed significantly to this subject
...
1
1
...
1
sin 3x
3
4
...
✂ cos 2 x ✂ e3 x
7
...
10
...
2 2
x ✞ 2 x2 ✞ 8 x ✞ C
7
14
...
x ✡ 3sin x + e ☛ C
7
3
3
...
4 3x
e ✁ x✁C
3
ax 3 bx 2
☎
☎ cx ☎ C 9
...
x2
4
✆ 5x ✆ ✆ C
x
2
13
...
6 2 4 2
x ✠ x ✠ 2x2 ✠ C
7
5
3
5
7
5
3
3
2
2 3
10
x ✞ 3cos x ✞ x 2 ✞ C
17
...
tan x – x + C
x
18
...
2 tan x – 3 sec x + C
22
...
C
EXERCISE 7
...
log (1 + x2) + C
2
...
cos (cos x) + C
5
...
2
4
( x ✠ 2) 2 ✟ ( x ✠ 2) 2 ✠ C
5
3
3
6
...
log 1+ log x ☛ C
1
cos 2( ax ✁ b) ✁ C
4a
5
3
ANSWERS
3
8
...
2
x ✄ 4( x ☎ 8) ✄ C
3
12
...
(log x)1✝ m
✞C
1✟ m
7
17
...
4 2
( x x 1) 2 C 10
...
✠
1
log | 9 ✠ 4 x 2 | 16
...
e tan
20
...
24
...
✆
C
✍
1
x
x ✁1 ✂ C
1
C
18(2 3x 3 ) 2
1 2 x✡3
e
☛C
2
x
✏x
19
...
1
tan (2 x ☎ 3) ☎ x ✄ C
2
☎ tan (7 ☎ 4 x) ✄ C
23
...
1
C
(1✆ tan x)
28
...
2sin x ✑ C
29
...
2
32
...
2 tan x ✔ C
37
...
B
1
4
27
...
C
– log (1+ cos x )
1
(1✄ log x)3 ✄ C
3
1
31
...
x 1
☎ log cos x ☎ sin x ✄ C
2 2
36
...
D
589
590
MATHEMATICS
EXERCISE 7
...
x 1
sin (4 x ✁ 10) ✁ C
2 8
3
...
4
...
1 ✠1
1
1
✡
cos 6 x ☛ cos 4 x ☛ cos 2 x✍ ☞ C
4 ✌✎ 6
4
2
✏
7
...
x
x tan ✁ C
2
11
...
1
1
cos 7 x ✁ cos x ✁ C
14
2
5
...
2 tan
x
x✁ C
2
3x 1
1
sin 2 x ✁ sin 4 x ✁ C
8 4
32
3x 1
1
✁ sin 4 x ✁ sin 8 x ✁ C
8 8
64
12
...
2 (sinx + x cos✒) + C
1
14
...
sec x – cosec x + C
1 3
tan x tan x ✁ x ✁ C
3
18
...
log tan x ✁ tan x ✁ C
2
20
...
21
...
A
2
22
...
1
cos ( x ✗ a )
log
✘C
sin (a ✗ b)
cos ( x ✗ b)
24
...
4
1
...
1
log 2 x ✁ 1 ✁ 4 x 2 ✁ C
2
ANSWERS
3
...
7
...
log tan x + tan x + 4 ✝ C
11
...
1
1 x3
log
6
1 ✁ x3
8
...
log x ✝ 1✝ x ✝ 2 x ✝ 2 ✝ C
–1
12
...
log x – ✟ x ✎ 3x ✟ 2 ✟ C
2
✑ 2 x ✏ 3✒
✕ ✓C
41 ✗
–1
14
...
2 2x 2 + x ✚ 3 ✛ C
18
...
9
2
2
19
...
C
✠ x ✟ 3✡
☞ ✟C
4 ✍
1 ✞1 ✠ 3 x ✟ 1✡
tan ☛
✟C
✌ 2 ☞✍
6
15
...
✫ x ✙ 2✬
✘C
✯ 2 ✮✰
– 4x – x2 ✘ 4 sin ✪1 ✭
21
...
x ✲1 ✲ 6
1
2
log x 2 ✲ 2 x ✲ 5 ✳
log
✳C
2
x ✲1✳ 6
6
591
592
MATHEMATICS
2
23
...
B
x2
4 x 10
C
25
...
5
1
...
x✄3
1
log
☎C
6
x☎3
3
...
1
3
log x ✞ 1 ✞ 2log x ✞ 2 ✟ log x ✞ 3 ✟ C
2
2
5
...
1
1
1
log x ✞ 1 ✞ log ( x 2 ✟ 1) ✟ tan ☛1 x ✟ C
2
4
2
8
...
11
...
9
...
– log x ✆ 1 + log (1 + x2) + tan–1x + C
2
12
...
3log x – 2 ✞
5
x✞2
16
...
x+
✟C
15
...
log
2
x
x
tan ✒1
✎ 3tan ✒1 ✏ C 19
...
4
x4
22
...
log ☎
23
...
6
1
...
✟ cos3 x ✠ sin 3 x ✠ C
3
9
3
...
x2
x2
log x ✡ ☛ C
2
4
5
...
x3
x3
log x ✡ ☛ C
3
9
7
...
x2
x 1
tan ✎1 x ✡ ☛ tan ✎1 x ☛ C
2
2 2
2
9
...
✏ sin –1 x✑ x ✁ 2 1 x2 sin✒1 x
2
2x✁C
11
...
x tan x + log cos x + C
13
...
✘
✂ x3 ✄
x3
15
...
ex sin x + C
17
...
e tan ✠ C
2
19
...
e2 x
(2sin x ✡ cos x) ☛ C
5
23
...
ex
✣C
( x ✤ 1) 2
22
...
B
593
594
MATHEMATICS
EXERCISE 7
...
1
x 4 ✁ x2
2
3
...
(x +2) 2
3
x ✂ 4 x ✂ 1 ✁ log x ✂ 2 ✂ x 2 ✂ 4 x ✂ 1
2
2
5
...
(x +2) 2
9
x ✂ 4 x ✁ 5 ✁ log x ✂ 2 ✂ x 2 ✂ 4 x ✁ 5
2
2
7
...
2x +3 2
x ☞ 3x
4
✂ 2sin
✌
1
x
✂C
2
2
...
A
✂C
✂C
✂C
☞C
✂C
9
...
D
EXERCISE 7
...
1 2 2
(b ✍ a )
2
2
...
27
2
5
...
19
3
6
...
64
3
EXERCISE 7
...
2
4
...
log
5
...
e4 (e – 1)
✂C
ANSWERS
7
...
1
log 2
2
8
...
15
...
5 –
✑
4
✒
✑
1024 2
20
...
☎
✝
11
...
17
...
595
✞
2
✞
4
5✡
5
3☛
✌ 9log ☞ log ✍
2✎
4
2✏
18
...
3log 2 ✔
21
...
C
3✓
8
EXERCISE 7
...
1
log 2
2
2
...
16 2
( 2 ✠ 1)
15
5
...
✞
64
231
✞
✞
e2 (e2 ✘ 2)
4
– log 2
2
1
6
...
8
10
...
17
log
21 ✗ 5 17
4
9
...
11
1
...
4
5
...
✞
8
✙
4
6
...
16 2
15
3
...
10
...
4
✙
4
1
(n ✚ 1) (n ✚ 2)
✞
2
log
1
2
11
...
✄
13
...
0
15
...
– ✄ log 2
17
...
5
20
...
C
MISCELLANEOUS EXERCISE ON CHAPTER 7
1
...
2 ( a ☛ x)
–
☞C
a
x
5
...
1
✌ 1 ✍4
4
...
✗ log x ✘ 1 ✘ log ( x ✘ 9) ✘ tan
2
4
2
3
7
...
✗ sin 2 x ✘ C
2
–1
9
...
8
...
sin (a – b)
cos( x ✦ a)
1 ✖1 4
sin ( x ) ✘ C
4
✧ 1+e x ★
C
13
...
1 ✖1 1 ✖ 1 x
tan x ✗ tan
✘C
3
6
2
1
4
15
...
1
log ( x 4 ✘ 1) ✘ C
4
18
...
19
...
–2 1– x ✁ cos
1
x ✁ x ✂ x2
21
...
✄ 2log
24
...
1✝
x cos ✆1 x ✄ 1 ✄ x 2 ✞✠ ☎ C
☛
2 ✡✟
✚
25
...
29
...
33
...
✛
6
4 2
3
✣
2
✄1
19
2
41
...
D
8
( 3 ✢ 1)
2
2sin ✜1
30
...
2✌
✏C
3 ✕✙
✛
28
...
✑
✣
2
(✣ ✄ 2)
1 ✤ 2 1✥
✧e ✦ ★
3✩
e✪
42
...
B
EXERCISE 8
...
14
3
4
...
10
...
16 ✫ 4 2
3
...
6✬
6
...
(4) 3
11
...
32 ✢ 8 2
3
✣
3
1
3
12
...
B
598
MATHEMATICS
EXERCISE 8
...
✁
9
sin
4
1
2 2
3
21
2
6
...
✄
✝
✝
✟
2
...
4
2✂
3
✆
3☎
✞
2 ✞✠
5
...
B
Miscellaneous Exercise on Chapter 8
1
...
8
2
...
6
...
27
9
...
7
16
...
7
3
4
...
9
2
5
...
2
9☞
8
18
...
C
14
...
12
...
B
EXERCISE 9
...
3
...
7
...
11
...
4
...
8
...
12
...
2
11
...
D
1
3
ANSWERS
EXERCISE 9
...
3
...
7
...
11
...
4
...
8
...
12
...
4
x
✁ x✂C
2
3
...
y 2 tan
2
...
tan x tan y ✄ C
5
...
tan –1 y = x +
7
...
x – 4 + y –4
9
...
y
x3
✝C
3
=C
10
...
✎✑ x 2 ✏✒
2
14
...
y – x + 2 = log (x2 (y + 2)2)
✔ y✓2 ✕ a
✘✖
x ✚
13
...
2y – 1 = ex ( sin x – cos x)
17
...
(x + 4) = y + 3
19
...
6
...
Rs 1648
2
22
...
A
EXERCISE 9
...
( x ★ y) ✩ Cx e
✧y
x
2
...
MATHEMATICS
y✁ 1
2
2
✆ ✂ log ( x ✄ y ) ✄ C
✝ x✞ 2
tan –1 ☎
1
5
...
xy cos
☞
✍
✏
x✟ 2y
x✡ 2y
✠
log x
✟C
y✌
✎ =C
x✑
4
...
y + x2 + y 2
8
...
cy = log ✥ 1
x
11
...
ye y
✦
x✧C
y
★
= ✩ log 2
2
x
y✌
✎ ✪ log ex
✏ x✑
☞
13
...
y + 2x = 3x2 y
y✌
✎ ✪ log ex
✏ x✑
☞
y✫
14
...
16
...
D
2x
( x ✬ 0, x ✬ e)
1 ✭ log x
EXERCISE 9
...
y =
3
...
y = (tan x – 1) + C e–tanx
✥2
4
...
y✮
8
...
y log x ✲
9
...
(x + y + 1) = C ey
11
...
x = 3y2 + Cy
✸
x
C
y
(1 ✩ log x ) ✩ C
2
...
y = cos x – 2 cos2 x
14
...
y = 4 sin3 x – 2 sin2 x
17
...
x + y + 1 = ex
18
...
D
Miscellaneous Exercise on Chapter 9
(ii) Order 1; Degree 3
1
...
y ✂✄
2 y 2 ✁ x2
4 xy
5
...
sin–1y + sin–1x = C
9
...
cos y =
☎
x
10
...
log x – y ✟ x ✠ y ✠ 1
13
...
31250
17
...
y e2
14
...
C
18
...
1
✔✔✔✕
1
...
602
MATHEMATICS
2
...
(i) scalar
(ii) vector (iii) scalar (iv) scalar
(v) scalar
(ii) scalar (iii) vector (iv) vector (v) scalar
✁
4
...
(i) True
(ii) False (iii) False (iv) False
EXERCISE 10
...
✆
✆
✆
a ☎ 3, b ☎ 62, c ☎ 1
2
...
3
...
4
...
– 7 and 6; –7 iˆ and 6 ˆj
6
...
1 ˆ 1 ˆ 2 ˆ
i✞
j✞
k
6
6
6
9
...
10
...
1
2
3
,
,
14 14 14
1 2 2
13
...
(i) ✠ iˆ ✡ ˆj ✡ kˆ (ii) ✝3iˆ ☛ 3 kˆ
3 3
3
16
...
(C)
19
...
3
1
...
8
...
cos –1 ✎ ✏
✑ 7✒
4
60
114
✕
6
...
16 2 2 2
,
3 7 3 7
13
3
...
6 a ☛ 11a
...
8
ANSWERS
12
...
✁3
2
✂
✄
14
...
cos ✝✟
10 ✆
✞
102 ✠
18
...
4
1
...
✡ iˆ ☛ ˆj ☛ kˆ
3
3
3
3
...
Either a ✌ 0 or b ✌ 0
2
8
...
3,
61
2
9
...
15 2
11
...
(C)
Miscellaneous Exercise on Chapter 10
1
...
3ˆ 1 ˆ
i✎ j
2
2
x2 – x1 , y2 – y1 , z2 ✏ z1; ( x2 ✏ x1 ) 2 ✑ ( y2 ✏ y1 ) 2 ✑ ( z2 ✏ z1 ) 2
✒5 ˆ
3 3ˆ
j
2
✂
✂
4
...
3
...
✔
i✓
1
3
8
...
3
10 ˆ
10 iˆ ✎
j
2
2
✂
9
...
✗ = 1
3
17
...
(C)
12
...
10
...
(B)
19
...
1
✂9 6 ✂ 2
1
1
1
, ,
,✁
,✁
3
...
0,
5
...
✁
EXERCISE 11
...
5
...
r ✄ (5 iˆ ✝ 4 ˆj ☎ 6 kˆ) ☎ ✆ (3 iˆ ☎ 7 ˆj ☎ 2 kˆ)
6
...
Vector equation of the line:
Cartesian equation of the line:
9
...
(i) ✌ = cos ✑
✕1 ✖ 8 ✗
(ii) ✌ = cos ✘✘
✙✙
✚5 3✛
✕1 ✖ 26 ✗
11
...
p✠
16
...
3 2
2
17
...
2 29
ANSWERS
EXERCISE 11
...
(a) 0, 0, 1; 2
2
3
,
,
14
14
(c)
2
...
(a) x + y – z = 2
(b) 2x + 3y – 4 z = 1
(c) (s – 2t) x + (3 – t) y + (2s + t) z = 15
☞ 24 36 48 ✌
,
,
✎
✏ 29 29 29 ✑
4
...
(a) [ r ✓
✗
(b) [ r ✓
(c) ✍
1 1✌
, ✎
3 3✑
☞
✏
18 24 ✌
,
✎
5
5 ✑
☞
✏
✒8
(b) ✍ 0,
(d) ✍ 0,
5
✌
✑
, 0✎
( iˆ ✓ 2 kˆ )] ✔ (iˆ ✕ ˆj ✓ kˆ) ✖ 0; x + y – z = 3
( iˆ ✕ 4 ˆj ✕ 6 kˆ ) ] ✔ ( iˆ ✓ 2 ˆj ✕ kˆ) ✖ 0; x – 2y + z + 1 = 0
6
...
There will be infinite number of planes
passing through the given points
...
10
...
y = 3
9
...
x – z + 2 = 0
✤ 15 ✥
✧
731 ✩
12
...
(a) cos✪1 ✍ ✎
✏ 5✑
(c) The planes are parallel
(e) 45o
3
13
(c) 3
14
...
90°
4
...
cos ☎
7
...
9
8
...
10
...
✒
13
...
(1, – 2, 7)
7
3
16
...
13
14
...
y – 3z + 6 = 0
17
...
r ✖ iˆ ✗ 2 ˆj ✗ 3 kˆ ✗ ✘ (✙ 3 iˆ ✗ 5 ˆj ✗ 4kˆ)
✚
20
...
B
22
...
1
1
...
Minimum Z = – 12 at (4, 0)
3
...
Minimum Z = 7 at ★✪ , ✩✫
2 2
5
...
Minimum Z = 6 at all the points on the line segment joining the points (6, 0)
and (0, 3)
...
Minimum Z = 300 at (60, 0);
Maximum Z = 600 at all the points on the line segment joining the points (120, 0)
and (60, 30)
...
Minimum Z = 100 at all the points on the line segment joining the points (0, 50)
and (20, 40);
Maximum Z = 400 at (0, 200)
9
...
No feasible region, hence no maximum value of Z
...
2
1
...
2☛
2
...
3
...
3 packages of nuts and 3 packages of bolts; Maximum profit = Rs 73
...
5
...
4 Pedestal lamps and 4 wooden shades; Maximum profit = Rs 32
7
...
8
...
9
...
100 kg of fertiliser F1 and 80 kg of fertiliser F2; Minimum cost = Rs 1000
11
...
40 packets of food P and 15 packets of food Q; Maximum amount of vitamin A
= 285 units
...
3 bags of brand P and 6 bags of brand Q; Minimum cost of the mixture = Rs 1950
3
...
608
MATHEMATICS
5
...
6
...
From A: 500, 3000 and 3500 litres; From B: 4000, 0, 0 litres to D, E and F
respectively; Minimum cost = Rs 4400
8
...
9
...
10
...
1
1
...
(i) 0
...
2
...
64
4
11
1
2
7
...
(i)
1
6
(ii)
16
25
(iii) 0
...
1
1 1
,
2 3
(ii)
1 2
,
2 3
12
...
0
(iii)
2
3
(iii)
6
7
10
...
(i)
14
...
(i)
8
...
1
1
, (b)
3
9
3 1
,
4 4
5
9
16
...
D
ANSWERS
609
EXERCISE 13
...
4
...
7
...
9
...
12
...
17
...
3
...
A and B are not independent
E and F are not independent
1
1
(i) p
(ii) p
5
10
(i) 0
...
58
(iii) 0
...
4
3
10
...
18
(ii) 0
...
72
(iv) 0
...
(i)
, (ii)
, (iii)
81
81
8
81
2
1
1
1
1
(i) , (ii)
15
...
(a) , (b) , (c)
3
2
5
3
2
D
18
...
3
1
2
198
5
...
9
13
...
9
8
10
...
C
1
...
52
5
11
...
3
...
4
11
12
...
EXERCISE 13
...
X = 0, 1, 2; yes
1
...
(i)
(ii)
X
0
1
2
P(X)
1
4
1
2
1
4
X
0
1
2
3
P(X)
1
8
3
8
3
8
1
8
3
...
(i)
(ii)
6
...
8
...
k
17
100
3
10
1
, P(X ☎ 2) 1, P(X ✆ 2)
2
1
14
10
...
5
11
...
3
3
13
...
833, S
...
415
9
...
53, Var(X) = 4
...
D(X) = 2
...
E(X) = 0
...
21
16
...
D
ANSWERS
EXERCISE 13
...
(i)
2
...
✂ ✄ ✂ ✄
☎ 20 ✆ ☎ 20 ✆
25
216
1
1024
5
...
95)5
(iv) 1 – (0
...
11
243
☞1✌
✑ ✒
✕ 2✖
7
...
(a) 1 ✗ ✂
7 5✁
11
...
C
5
243
1024
(iii) 1 – (0
...
2
(ii)
4
(b)
63
64
9
45
512
(ii) (0
...
2
4
...
(iii)
(iii)
20
20
20
✓✍ 20C12 ✏ C13 ✏
...
✂ ✄
18 ☎ 6 ✆
15
...
149 99 ✁
100 ✂☎ 100 ✄✆
22 ✘ 93
1011
Miscellaneous Exercise on Chapter 13
1
...
(i)
3
...
1 ✛ ✜ 10 Cr (0
...
1)10✙ r
r✚7
5
...
510
2 69
9
...
7
...
A
8
...
(i) 0
...
n ✟ 4
13
...
11
216
14
...
B
16
...
05
18